Answer:
The test statistic for the hypothesis test is -1.202.
Step-by-step explanation:
We are given that a report on the nightly news broadcast stated that 10 out of 129 households with pet dogs were burglarized and 23 out of 197 without pet dogs were burglarized.
Let [tex]p_1[/tex] = population proportion of households with pet dogs who were burglarized.
[tex]p_2[/tex] = population proportion of households without pet dogs who were burglarized.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]p_1=p_2[/tex] {means that both population proportions are equal}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]p_1\neq p_2[/tex] {means that both population proportions are not equal}
The test statistics that would be used here Two-sample z-test for proportions;
T.S. = [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ~ N(0,1)
where, [tex]\hat p_1[/tex] = sample proportion of households with pet dogs who were burglarized = [tex]\frac{10}{129}[/tex] = 0.08
[tex]\hat p_2[/tex] = sample proportion of households without pet dogs who were burglarized = [tex]\frac{23}{197}[/tex] = 0.12
[tex]n_1[/tex] = sample of households with pet dogs = 129
[tex]n_2[/tex] = sample of households without pet dogs = 197
So, the test statistics = [tex]\frac{(0.08-0.12)-(0)}{\sqrt{\frac{0.08(1-0.08)}{129}+\frac{0.12(1-0.12)}{197} } }[/tex]
= -1.202
The value of z test statistics is -1.202.
Evaluate: 5-^2 =
pls help
Answer:
1/25
Step-by-step explanation:
5^-2
We know that a^ -b = 1/ a^b
5^-2 = 1/ 5^2
= 1/25
Alguien me puede ayudar con en esto por favor !!!
Answer:
y=cosx
Step-by-step explanation:
cosx has a domain of all real numbers
This Question: 4 pts
1 of 11 (0 complete)
Music Preferences
Students at a high school were polled to determine the type of music they preferred. There were 1960 students who
completed the poll. Their responses are represented in the circle graph.
Rap 916
Alternative 46
Rock and Roll 279
Country 183
Jazz 32
Other 89
About What % of the students who completed the poll preferred rock and roll music.
(Round to one decimal place as needed.)
Answer:
The percentage of the students who completed the poll preferred rock and roll music.
P(RR) = 0.1423 = 14.23 %
Step-by-step explanation:
Explanation:-
Given total number of students n(S) = 1960
Given the Students at a high school were polled to determine the type of music they preferred.
Rap 916
Alternative 46
Rock and Roll 279
Country 183
Jazz 32
Other 89
Let ' RR' be the event of Rock and Roll preferred music
given Rock and Roll = 279
n( RR) = 279
The percentage of the students who completed the poll preferred rock and roll music.
[tex]P(RR) = \frac{n(RR)}{n(s)} = \frac{279}{1960}[/tex]
P(RR) = 0.1423 = 14.23 %
Which point is coplanar with B , C , H ?
Answer:
G
Step-by-step explanation:
Point G is coplanar with points B, C, H.
A recent graduate school study of a random sample of 250 US manufacturing companies determined the average financial report preparation time was 68.04 days with a standard deviation of 35.74 days. Calculate to three decimal places the 95 percent confidence interval for the mean report prep time for all US manufacturing companies. [63.001, 72.008] [63.957, 75.568] [63.505, 72.414] [61.612, 74.468] [63.612, 72.468]
Answer:
[tex]68.04-1.959\frac{35.74}{\sqrt{250}}=63.618[/tex]
[tex]68.04+1.959\frac{35.74}{\sqrt{250}}=72.461[/tex]
And the best option for this case would be
Step-by-step explanation:
Information given
[tex]\bar X=68.04[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s=35.74 represent the sample standard deviation
n=250 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=250-1=249[/tex]
The Confidence level is 0.95 or 95%, and the significance [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value for this case woud be [tex]t_{\alpha/2}=1.956[/tex]
And replacing we got:
[tex]68.04-1.959\frac{35.74}{\sqrt{250}}=63.618[/tex]
[tex]68.04+1.959\frac{35.74}{\sqrt{250}}=72.461[/tex]
And the best option for this case would be
What’s the correct answer for this question?
Answer:
B.
Step-by-step explanation:
A cone with a radius of 1-unit will be obtained by rotating the 3-D figure.
SOMEONE PLEASE HELP ME ASAP PLEASE!!!
Answer:
105.12 ft^2
Step-by-step explanation:
Area of a rectangle: bh
In this case 8*10.... so area of the rectangle is 80
Area of a circle: pir^2
Half it for a semicircle.
so 1/2 pi r^2
radius is 4 cuz its half of 8.
so 1/2(3.14)(4^2)=(0.5)(3.14)(16)=25.12
Now add up 80+25.12
Total is 105.12
Hope I helped :)
You need a 55% alcohol solution. On hand, you have a 525 mL of a 45% alcohol mixture. You also have 90% alcohol mixture. How much of the 90% mixture will you need to add to obtain the desired solution? You will need _____mL of the 90% solution to obtain _____mL of the desired 55% solution.
Answer:
Let's call the amount of 90% solution x.
We can write:
45% * 525 + 90%x = 55%(x + 525)
0.45 * 525 + 0.9x = 0.55(x + 525)
Solving for x we get x = 150 so the first blank is 150 and the second blank is 525 + 150 = 675.
what is the value of this expression plssssss 8z-3 when z =7
Answer:
53
Step-by-step explanation:
8•7 is 56
56 - 3 is 53
Answer:
53
Step-by-step explanation:
z = 7
8z is the same as saying 8×z
8×7-3 (do multiplication first)
56-3 = 53
A multiple-choice standard test contains total of 25 questions, each with four answers. Assume that a student just guesses on each question and all questions are answered independently. (a) What is the probability that the student answers more than 20 questions correctly
Answer:
[tex]P(x>20)=9.67*10^{-10}[/tex]
Step-by-step explanation:
If we call x the number of correct answers, we can said that P(x) follows a Binomial distribution, because we have 25 questions that are identical and independent events with a probability of 1/4 to success and a probability of 3/4 to fail.
So, the probability can be calculated as:
[tex]P(x)=nCx*p^{x}*q^{n-x}=25Cx*0.25^{x}*0.75^{25-x}[/tex]
Where n is 25 questions, p is the probability to success or 0.25 and q is the probability to fail or 0.75.
Additionally, [tex]25Cx=\frac{25!}{x!(25-x)!}[/tex]
So, the probability that the student answers more than 20 questions correctly is equal to:
[tex]P(x>20)=P(21)+P(22)+P(23)+P(24)+P(25)[/tex]
Where, for example, P(21) is equal to:
[tex]P(21)=25C21*0.25^{21}*0.75^{25-21}=9.1*10^{-10}[/tex]
Finally, P(x>20) is equal to:
[tex]P(x>20)=9.67*10^{-10}[/tex]
the length of a ruler is 170cm,if the ruler broke into four equal parts.what will be the sum of the length of three parts
Answer:
Step-by-step explanation:Srry it's bit rough...
Ruby has a bird feeder which is visited by an average of 13 birds every 2 hours during daylight hours. What is the probability that the bird feeder will be visited by more than 3 birds in a 40 minute period during daylight hours? Round your answer to three decimal places.
Answer:
62.93%
Step-by-step explanation:
We have to solve it by a Poisson distribution, where:
p (x = n) = e ^ (- l) * l ^ (x) / x!
Where he would come being the number of birds that there would be in 40 minutes, we know that in 2 hours, that is 120 minutes there are 13, therefore in 40 there would be:
l = 13 * 40/120
l = 4,333
Now, we have p (x> 3) and that is equal to:
p (x> 3) = 1 - p (x <= 3)
So, we calculate the probability from 0 to 3:
p (x = 0) = 2.72 ^ (- 4.33) * 4.33 ^ (0) / 0! = 0.01313
p (x = 1) = 2.72 ^ (- 4.33) * 4.33 ^ (1) / 1! = 0.0568
p (x = 2) = 2.72 ^ (- 4.33) * 4.33 ^ (2) / 2! = 0.12310
p (x = 3) = 2.72 ^ (- 4.33) * 4.33 ^ (3) / 3! = 0.17767
If we add each one:
0.01313 + 0.0568 + 0.12310 + 0.17767 = 0.3707
replacing:
p (x> 3) = 1 - 0.3707
p (x> 3) = 0.6293
Which means that the probability is 62.93%
PLEASE HELLLLP!!!! If x+y−z=8 and x−y+z=12, then x=
Answer:
(C) 10
Step-by-step explanation:
x+y−z=8
Subtract y from both sides
x-z= -y+8
Add z to both sides
x= -y+z+8
Subtract 8 from both sides
x-8= -y+z
x−y+z=12
Add y to both sides
x+z=12+y
Subtract z from both sides
x=y-z+12
Subtract 12 from both sides
x-12=y-z
Multiply both sides by -1
-x+12= -y+z
Combine equations:
x-8= -x+12
Add x to both sides
2x-8+12
Add 8 to both sides
2x=20
Divide both sides by 2
x=10
The answer is (C) 10.
Among fatal plane crashes that occurred during the past 65 years, 627 were due to pilot error, 64 were due to other human error, 113 were due to weather, 382 were due to mechanical problems, and 481 were due to sabotage. Construct the relative frequency distribution.
What is the most serious threat to aviation safety, and can anything be done about it?
A. Sabotage is the most serious threat to aviation safety. Airport security could be increased.
B. Mechanical problems are the most serious threat to aviation safety. New planes could be better engineered.
C. Weather is the most serious threat to aviation safety. Weather monitoring systems could be improved.
D. Pilot error isPilot error is the most serious threat to aviation safety. Pilots could be better trained.
Answer:
D. Pilot error isPilot error is the most serious threat to aviation safety. Pilots could be better trained.
Step-by-step explanation:
We can construct the relative frequency distribution dividing the amount of incidents for each category by the total amount of incidents.
This amount is:
[tex]\sum x_i=627+64+113+382+481=1667[/tex]
Then, the relative frequency for each category is:
[tex]\text{Pilot error}=627/1667=0.38\\\\\text{Human error}=64/1667=0.04\\\\\text{Weather}=113/1667=0.07\\\\\text{Mechanical problems}=382/1667=0.23\\\\\text{Sabotage}=481/1667=0.29\\\\[/tex]
As the pilot error has the largest relative frequency, we can conclude that pilot error is the most serious threat to aviation safety.
What value of x makes 3(x + 4) = 3x + 4 true?
Well lets see.
[tex]3(x+4)=3x+4\implies 12 = 4\implies x\notin\mathbb{C}[/tex].
There are no such x-es that satisfy the equation.
In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 69.9 inches and a standard deviation of 4.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below. (a) Find the probability that a study participant has a height that is less than 65 inches. The probability that the study participant selected at random is less than 65 inches tall is nothing. (Round to four decimal places as needed.)
Answer:
The probability that a study participant has a height that is less than 65 inches is 0.1103.
Step-by-step explanation:
We are given that the heights in the 20-29 age group were normally distributed, with a mean of 69.9 inches and a standard deviation of 4.0 inches.
A study participant is randomly selected.
Let X = heights in the 20-29 age group.
So, X ~ Normal([tex](\mu=69.9,\sigma^{2} =4.0^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean height = 69.9 inches
[tex]\sigma[/tex] = standard deviation = 4.0 inches
Now, the probability that a study participant has a height that is less than 65 inches is given by = P(X < 65 inches)
P(X < 65 inches) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{65-69.9}{4}[/tex] ) = P(Z < -1.225) = P(Z [tex]\leq[/tex] 1.225)
= 1 - 0.8897 = 0.1103
The above probability is calculated by looking at the value of x = 1.225 in the z table which lies between x = 1.22 and x = 1.23 which has an area of 0.88877 and 0.89065 respectively.
123 grams is rounded to nearest whole. Write down the minimum possible mass it could have been.
Answer:
The nearest whole is 122.99 repeated
Step-by-step explanation:
The accompanying data are the times to failure (in millions per cycle) of high-speed turbine engine bearings made out of two different compounds. These were taken from "Analysis of Single Classification Experiments Based on Censored Samples from the Two-parameter Weibull Distribution" by J.I. McCool (The Journal of Statistical Planning and Inference, 1979) Compound 1 3.03 5.53 5.60 9.30 9.92 12.51 12.95 15.21 16.04 16.84 Compound 2 3.19 4.26 4.47 4.53 4.67 4.69 5.78 6.79 9.37 12.75 (a) Find the 0.84 quantile of the Compound 1 failure times (b) Give the coordinates of the two lower-left points that would appear on a normal plot of the compound 1 data (c) Make back-to-back stem-and-leaf plots for comparing the life length properties of bearings made from Compounds 1 and 2 (d) Make (to scale) side-by-side boxplots for comparing the life lengths for the two compounds. Mark numbers on the plots indicating the locations of their main features (e) Compute the sample means and standard deviations of the two sets of lifetimes (f) Describe what your answers to parts (c), (d), and (e) above indicate about the life lengths of these turbine bearings.
Find the given attachments
The total mass of the Sun is about 2×10^30 kg, of which about 76 % was hydrogen when the Sun formed. However, only about 12 % of this hydrogen ever becomes available for fusion in the core. The rest remains in layers of the Sun where the temperature is too low for fusion.
Part A
Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the Sun.
Express your answer using two significant figures.
Part B
The Sun fuses about 600 billion kilograms of hydrogen each second. Based on your result from part A, calculate how long the Sun’s initial supply of hydrogen can last. Give your answer in both seconds and years.
Express your answer using two significant figures.
Part D
Given that our solar system is now about 4.6 billion years old, when will we need to worry about the Sun running out of hydrogen for fusion?
Express your answer using two significant figures.
Answer:
A. 1.8 ×[tex]10^{30}[/tex] Kg
B i. 3.0 × [tex]10^{17}[/tex] seconds
ii. 9.6 × [tex]10^{9}[/tex] years
C. After 9.2 × [tex]10^{9}[/tex] (9.2 billion) years
Step-by-step explanation:
Given that the mass of the Sun = 2× [tex]10^{30}[/tex] Kg.
Mass of hydrogen when Sun was formed = 76% of 2× [tex]10^{30}[/tex] Kg
= [tex]\frac{76}{100}[/tex] ×2× [tex]10^{30}[/tex] Kg
= 1.52 × [tex]10^{30}[/tex] Kg
Mass of hydrogen available for fusion = 12% of 1.52 × [tex]10^{30}[/tex] Kg
= [tex]\frac{12}{100}[/tex] × 1.52 × [tex]10^{30}[/tex] Kg
= 1.824 ×[tex]10^{30}[/tex] Kg
A. Total mass of hydrogen available for fusion over the lifetime of the sun is 1.8 ×[tex]10^{30}[/tex] Kg.
B. Given that the Sun fuses 6 × [tex]10^{11}[/tex] Kg of hydrogen each second.
i. The Sun's initial hydrogen would last;
[tex]\frac{1.8*10^{30} }{6*10^{11} }[/tex]
= 3.04 × [tex]10^{17}[/tex] seconds
The Sun's hydrogen would last 3.0 × [tex]10^{17}[/tex] seconds
ii. Since there are 31536000 seconds in a year, then;
The Sun's initial hydrogen would last;
[tex]\frac{3.04*10^{17} }{31536000}[/tex]
= 9.640 × [tex]10^{9}[/tex] years
The Sun's hydrogen would last 9.6 × [tex]10^{9}[/tex] years.
C. Given that our solar system is now about 4.6 × [tex]10^{9}[/tex] years, then;
[tex]\frac{9.6*10^{9} }{4.6*10^{9} }[/tex]
= 2.09
So that; 2 × 4.6 × [tex]10^{9}[/tex] = 9.2 × [tex]10^{9}[/tex] years
Therefore, we need to worry about the Sun running out of hydrogen for fusion after 9.2 × [tex]10^{9}[/tex] years.
Part(A): The total mass of hydrogen available 9.6 billion years.
Part(B): The total time is 5.10 billion years.
Part(D): The hydrogen will last [tex]5.04\times 10^9 \ years[/tex]
Mass of the sun:Our sun is the largest object in our solar system. The mass of the sun is approximately [tex]1.988\times 1030[/tex] kilograms
Part(A):
Given that,
The total mass of the Sun =[tex]2\times10^{30} kg[/tex]
Mass of hydrogen in Sun = [tex]2\times10^{30} \times0.76\ kg[/tex]
The mass of hydrogen ever available for fusion is,
[tex]2\times10^{30} \times 0.76 \times 0.12 kg = 1.824\times 10^{29}[/tex]
Mass of hydrogen fuses each second = 600 billion kg/second.
Time hydrogen will last in seconds=[tex]1.824\times 10^{29}seconds.[/tex]
[tex]= 0.00304\times 10^{29 }= 3.04\times 10^{17}.[/tex]
Time hydrogen will last in seconds =[tex]1 year = 31,536,000 seconds.[/tex]
[tex]31,536,000 x = 3.04\times 10^{17} = 9.6 \ billion \ years.[/tex]
(B) Present age of sun = [tex]9.6-4.5 \ billion \ years[/tex]
The time when we need to worry about Sun running out of hydrogen for fusion = 5.10 billion years.
(D) The solar system is 4.6 billion years old that is [tex]4.6\times 10^9\ years[/tex]
And in part (B) we have calculated that hydrogen will last [tex]9.64\times 10^9[/tex] then,
[tex]9.64\times 10^9-4.6\times 10^9=5.04\times 10^9[/tex]
Learn more about the topic mass of the sun:
https://brainly.com/question/11359658
A phone charger requires 0.5 A at 5V. It is connected to a transformer with 100 % of efficiency whose primary contains 2200 turns and is connected to 220-V household outlet.
(a) How many turns should there be in the secondary?
(b) What is the current in the primary?
(c) What would be the output current and output voltage values if number of secondary turns (N2) doubled of its initial value?
Answer:
a. 50 turns
b. 0.0114 A
c. 0.25 A, 10 V
Step-by-step explanation:
Given:-
- The required current ( Is ) = 0.5 A
- The required voltage ( Vs ) = 5 V
- Transformer is 100% efficient ( ideal )
- The number of turns in the primary coil, ( Np ) = 2200
- The Voltage generated by power station, ( Vp ) = 220 V
Find:-
a. The number of turns in the secondary coil of the transformer
b. The current supplied by the power station
c. The effect on output current and voltage when the number of turns of secondary coil are doubled.
Solution:-
- For ideal transformers that consists of a ferromagnetic core with two ends wounded by a conductive wire i.e primary and secondary.
- The power generated at the stations is sent to home via power lines and step-down before the enter our homes.
- A household receives a voltage of 220 V at one of it outlets. We are to charge a phone that requires 0.5 A and 5V for the process.
- The outlet and any electronic device is in junction with a smaller transformer.
- All transformers have two transformation ratios for current ( I ) and voltage ( V ) that is related to the ratio of number of turns in the primary and secondary.
Voltage Transformation = [tex]\frac{N_p}{N_s} = \frac{V_p}{V_s}[/tex]
Where,
Ns : The number of turns in secondary winding
- Plug in the values and evaluate ( Ns ):
[tex]N_s = N_p*\frac{V_s}{V_p} \\\\N_s = 2200*\frac{5}{220} \\\\N_s = 50[/tex]
Answer a: The number of turns in the secondary coil should be Ns = 50 turns.
- Similarly, the current transformation is related to the inverse relation to the number of turns in the respective coil.
Current Transformation = [tex]\frac{N_p}{N_s} = \frac{I_s}{I_p}[/tex]
Where,
Ip : The current in primary coil
- Plug in the values and evaluate ( Ip ):
[tex]I_p = \frac{N_s}{N_p}*I_s\\\\I_p = \frac{50}{2200}*0.5\\\\I_p = 0.0114[/tex]
Answer b: The current in the primary coil should be Ip = 0.0114 Amp.
- The number of turns in the secondary coil are doubled . From the transformation ratios we know that that voltage is proportional to the number of turns in the respective coils. So if the turns in the secondary are doubled then the output voltage is also doubled ( assuming all other design parameters remains the same ). Hence, the output voltage is = 2*5V = 10 V
- Similary, current transformation ratio suggests that the current is inversely proportional to the number of turns in the respective coils. So if the turns in the secondary are doubled then the output current is half of the required ( assuming all other design parameters remains the same ). Hence, the output current is = 0.5*0.5 A = 0.25 A
85% of z is 106,250. What is z?
Answer:
z=12500
Step-by-step explanation:
Of means multiply and is means equals
85% *z = 106250
Change to decimal form
.85z = 106250
Divide each side by .85
.85z/.85 = 106250 /.85
z=12500
How many solutions does the system have? y = -2x-4 \\\\ y = 3x+3
Answer:
The system has one solution.
Step-by-step explanation:
We have two equations:
y = -2x - 4
y = 3x + 3
Equalling them:
y = y
-2x - 4 = 3x + 3
5x = -7
[tex]x = -\frac{7}{5}[/tex]
And
[tex]y = 3x + 3 = 3(-\frac{7}{5}) + 3 = \frac{-21}{5} + 3 = \frac{-21}{5} + \frac{15}{5} = -\frac{6}{5}[/tex]
Replacing in the other equation we should get the same result.
[tex]y = -2x - 4 = -2(-\frac{7}{5}) - 4 = \frac{14}[5} - 4 = \frac{14}{4} - \frac{20}{5} = -\frac{6}{5}[/tex]
So the system has one solution.
What’s the correct answer for this question?
Answer:
C.
Step-by-step explanation:
First finding height using Pythagoras theorem
(H)²=(B)²+(P)²
8.2²=5.4²+P²
P² = 67.24 - 29.16
P² = 38.08
P = 6.2
Now
Volume of cone = (1/3)πr²h
= (1/3)(3.14)(5.4)²(6.2)
= (1/3)(567.9)
= 189.2 cm³
Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math has 40 students and earns the college $40,000 in revenue. Each section of Applied Calculus has 40 students and earns the college $60,000, while each section of Computer Methods has 10 students and earns the college $26,000. Assuming the college wishes to offer a total of seven sections, accommodate 220 students, and bring in $292,000 in revenues, how many sections of each course should it offer?
Finite Math section(s)
Applied Calculus section(s)
Computer Methods section(s)
Answer:
meh
Step-by-step explanation:
A boy is playing a ball in a garden surrounded by a wall 2.5 m high and kicks the ball vertically up from a height of 0.4 m with a speed of 14 m/s. For how long is the ball above
the height of the wall.
Answer:
2.54 seconds
Step-by-step explanation:
We can use the following equation to model the vertical position of the ball:
S = So + Vo*t + a*t^2/2
Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.
Then, using S = 2.5, So = 0.4, Vo = 14 and a = -9.8 m/s2, we have that:
2.5 = 0.4 + 14*t - 4.9t^2
4.9t^2 - 14t + 2.1 = 0
Solving this quadratic equation, we have that t1 = 2.6983 s and t2 = 0.1588 s.
Between these times, the ball will be higher than 2.5 m, so the amount of time the ball will be higher than 2.5 m is:
t1 - t2 = 2.6983 - 0.1588 = 2.54 seconds
Which triangle congruence postulate can be used to prove that FEH=HGF?
SAS
HL
ASA
SSS
Answer:
Option (2).
Step-by-step explanation:
From the figure attached,
EFGH is a quadrilateral and FH is line which divides the quadrilateral into two right triangles, ΔFEH and ΔHGF.
In ΔFEH and ΔHGF,
Sides EH ≅ FG [Given]
FH ≅ FH [reflexive property]
ΔFEH ≅ ΔHGF [HL (Hypotenuse - length) postulate of congruence]
Option (2) will be the answer.
You wake up one morning, and find yourself wearing a toga and scarab ring. Always a logical person, you conclude that you must have become an Egyptian pharoah. You decide to honor yourself with a pyramid of your own design. You decide it should have height h = 130 and a square base with side s = 870
To impress your Egyptian subjects, find the volume of the pyramid.
Answer:
32799000 cubic units
Step-by-step Explanation:
Height if the Pyramid=130 Units
Side Length of square base=870 Units
Volume of a Pyramid=[tex]\frac{1}{3}[/tex]* Base Area*Height
Since the base is a square,
Area of a Square of side length s[tex]=s^2[/tex]
Therefore:
Volume[tex]=\frac{1}{3}*870^2*130[/tex]
=32799000 cubic units
The volume of your pyramid is 32799000 cubic units.
Answer:
V = 3.28x10⁷
Step-by-step explanation:
The volume of a pyramid is given by:
[tex] V = \frac{1}{3}bh [/tex]
Where:
b: is the base of the pyramid
h: is the height of the pyramid = 130
The base of the square base of the pyramid is given by:
[tex] b = s^{2} [/tex]
Where:
s: is the side of the square base = 870
Thus, the base of the square base of the pyramid is:
[tex] b = s^{2} = (870)^{2} = 7.56 \cdot 10^{5} [/tex]
Now, the volume of a pyramid is:
[tex] V = \frac{1}{3}bh = \frac{1}{3}(7.56 \cdot 10^{5})*130 = 3.28 \cdot 10^{7} [/tex]
Therefore, the volume of the pyramid is 3.28x10⁷.
I hope it helps you!
Write the number one and three quarter million in figures.
Answer 1750000
Step-by-step explanation:
The number "one and three-quarter million" can be written in figures as 1,750,000.
Given the number one and three quarter million.
"One" represents the digit 1.
"Three-quarter" means three-fourths or 3/4.
In decimal form, this fraction is 0.75.
"Million" signifies a value of 1,000,000.
To represent this number in figures, we combine these components:
We start with the digit 1, representing "one."
Then we append the value 750,000, which corresponds to "three-quarter million."
Combining these two parts, we have 1,750,000.
Therefore, "one and three-quarter million" can be written in figures as 1,750,000.
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5 of 5
It is worked out that if 5 ladles full of soup are given to
each person,
140 people can be fed.
The customers have complained in the past that the
portions are too small.
The cook decides to give 7 ladles full of soup to each
person.
How many people can now be fed soup?
people
Answer:
Number of people that can be served 7 ladles = 100 people
Step-by-step explanation:
We are told that;
Initial number of ladles proposed per person = 5
Number of persons to be fed based on 5 ladles = 140 persons
Thus, amount of ladles based on that data is;
140 people x 5 ladle/1 person = 700 ladles full of soup
Now, since the cook decides to give 7 ladles full of soup to each person, the number of people that can be fed will now be;
700 ladles ÷ 7 ladles/person = 100 persons
Salinas Corporation has net income of $15 million per year on net sales of $90 million per year. It currently has no long-term debt but is considering a debt issue of $20 million. The interest rate on the debt would be 7%. Salinas Corp. currently faces an effective tax rate of 40%. What would be the annual interest tax shield to Salinas Corp. if it goes through with the debt issuance?
Answer:
The annual interest tax shield to Salinas Corp would be of $560,000
Step-by-step explanation:
In order to calculate the annual interest tax shield to Salinas Corp if it goes through with the debt issuance we would have to calculate the following formula:
Annual Interest tax shield = Interest * tax
Interest = debt *rate of interest
Interest=$20 million * 0.07
Interest= $ 1.40 million
tax= 40%
Therefore, Annual Interest tax shield =$1.40 million * 0.40
Annual Interest tax shield = $560,000
The annual interest tax shield to Salinas Corp would be of $560,000
