Answer:
14.28 mm
Step-by-step explanation:
Find the circumference if it were a normal circle, then divide it by 4.
C = 2[tex]\pi[/tex]r
C = 2[tex]\pi[/tex](4)
C = 8[tex]\pi[/tex]
Divide it by 4
2[tex]\pi[/tex] + 4 + 4 = 14.28
Answer:
25.13 mm is the circumfrence, I believe.. Been a while since I've worked with this
Step-by-step explanation:
Suppose that y = 5 x plus 4 and it is required that y be within 0.005 units of 8. For what values of x will this be true?
Answer:
so we have an inequality for y -
7.995<y<8.005
Then now we need in inequality for x
(y-4)/5 = x
so that means that so if we have (7.995-4)/5 we get 3.995/5 = 7.99
so we have our first 7.99<x<b
Now we solve for b
So that means that 5.005/5 = 1.001
since we are changing it we switch our signs
from 7.99<x<1.001
we do 7.99>x>1.001
therefore
1.001<x<7.99
Answer:
0.795 [tex]\leq[/tex] y [tex]\leq[/tex] 0.805
Step-by-step explanation:
8 = 5x + 4
5x = 4
x = 4/5 or 0.800
therefore 0.800 + .005 and 0.800 - .005 =
0.795 [tex]\leq[/tex] y [tex]\leq[/tex] 0.805
How do u solve this?
Answer:
0
Step-by-step explanation:
Tuesday : -1/2
Wednesday + 3/4
Thursday : -3/8
Add them together
-1/2 + 3/4- 3/8
Get a common denominator
-4/8 + 6/8 - 3/8
-1/8
The closest integer value to -1/8 is 0
HELP ME QUICK!! The best answer I will mark brainlest!
Answer: 1. (4, 8); 2. (3, 4)
Step-by-step explanation: I tried to get this to you fast but I can give you an explaination if you would like one :)
What’s the correct answer for this question?
Answer:
A: 97π/18 m
Step-by-step explanation:
Central Angle = 97°
In radians:
97° = 97π/180
Now
S = r∅
S = (10)(97π/180)
S = 97π/18 m
Answer:
The answer is option 1.
Step-by-step explanation:
Given that the formula for length of Arc is Arc = θ/360×2×π×r when r represents the radius of circle. Then, you have to substitute the following values into the formula :
[tex]arc = \frac{θ}{360} \times 2 \times \pi \times r[/tex]
Let θ = ∠VCW = 97°,
Let r = 10m,
[tex]arc = \frac{97}{360} \times 2 \times \pi \times 10[/tex]
[tex]arc = \frac{97}{360} \times 20 \times \pi[/tex]
[tex]arc = \frac{97}{18} \pi \: m[/tex]
A bag contains 1p,20 and 5p coins 3/8 of the bag are 1p coins There are as many 5p coins as 1p coins in the bag. There are 640 coins in total. Work out the number of 20 coins in the bag
Answer:
160 off 20p coins
Step-by-step explanation:
1 p, 20 p, 5 p coins1 p= 3/8 of the bag5 p= 1 p= 3/8 of the bagtotal coins= 64020 p coins= 640 - 640*(3/8+3/8)= 640*(1- 6/8)= 640 * 2/8= 640* 1/4= 160
The percent, X , of shrinkage o n drying for a certain type of plastic clay has an average shrinkage percentage :, where parameter : is unknown. A random sample of 45 specimens from this clay showed an average shrinking percentage of 18.4 and a standard deviation of 2.2.
Required:
a. Estimate at 5% level of significance whether the true average shrinkage percentage U: is greater than 17.5 and write your conclusion.
b. Report the p-value.
Answer:
a) [tex]t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744[/tex]
The degrees of freedom are given by:
[tex]df=n-1=45-1=44[/tex]
The critical value for this case is [tex]t_{\alpha}=1.68[/tex] since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4
b) [tex]p_v =P(t_{(44)}>2.744)=0.0044[/tex]
Step-by-step explanation:
Information given
[tex]\bar X=18.4[/tex] represent the sample mean
[tex]s=2.2[/tex] represent the sample standard deviation
[tex]n=45[/tex] sample size
[tex]\mu_o =17.5[/tex] represent the value to verify
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
Part a
We want to test if the true mean is higher than 17.5, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 17.5[/tex]
Alternative hypothesis:[tex]\mu > 17.5[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
And replacing we got:
[tex]t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744[/tex]
The degrees of freedom are given by:
[tex]df=n-1=45-1=44[/tex]
The critical value for this case is [tex]t_{\alpha}=1.68[/tex] since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4
Part b
The p value would be given by:
[tex]p_v =P(t_{(44)}>2.744)=0.0044[/tex]
What is a word problem for 15 minus 28?
Answer:
A word problem for that would be Sam had 28 chocolates and Bob took away 15. How many does Sam have left?
Step-by-step explanation:
I don't know how to show work for writing a word problem. Sorry
Answer:
Step-by-step explanation:
Jane has $15 in her bank account. She wrote a $28 check for buying a fiction book. How much is her balance now?
please help me explain your answer only answer if you are sure
Answer:
The answer of top prism is 262
and down prism is 478
The upper figure is triangular prism.
so, we use bh+2ls+lb formula
B=5
h=3
s=4
l=19
Now,
surface area of triangular prism = bh+2ls+lb
= 5×3+2×19×4+19×5
= 262
The down figure is rectangular prism.
so, we use 2lw+2lh+2hw
l=5
h=6
w=19
Now,
The area of rectangular prism = 2lw+2lh+2hw
= 2×5×19+2×19×6+2×5×6
= 478
Henry, Brian and Colin share some sweets in the ratio 6:4:1. Henry gets 25 more sweets than Colin. How many sweets are there altogether?
Answer:
There are 55 sweets in total.
Step-by-step explanation:
The total number of sweets is t.
Henry, Brian and Colin share some sweets in the ratio 6:4:1.
This means that Henry earns [tex]\frac{6}{6+4+1} = \frac{6}{11}[/tex] of the total(t).
Brian earns [tex]\frac{4}{11}[/tex] of the total.
Colin earns [tex]\frac{1}{11}[/tex] of the total
Henry gets 25 more sweets than Colin.
Henry earns [tex]\frac{6t}{11}[/tex]
Colin earns [tex]\frac{t}{11}[/tex]
So
[tex]\frac{6t}{11} = \frac{t}{11} + 25[/tex]
Multiplying everything by 1
[tex]6t = t + 275[/tex]
[tex]5t = 275[/tex]
[tex]t = \frac{275}{5}[/tex]
[tex]t = 55[/tex]
There are 55 sweets in total.
Use the table of values to find the line of regression and if justified at the 0.05 significance level, use it to find the predicted quality score of a TV set with a price of $1900. If the data does not suggest linear correlation, then use the average quality score as a prediction.
Price: 2,300, 1,800, 2,500, 2,700, 2,000, 1,700, 1,500, 2,700
Quality Score: 74, 73, 70, 66, 63, 62, 52, 68
Answer:
Step-by-step explanation:
no x y xy x²
1 2300 74 170200 5290000
2 1800 73 131400 3240000
3 2500 70 175000 6250000
4 2700 66 178200 7290000
5 2000 63 126000 4000000
6 1700 62 105400 2890000
7 1500 52 78000 2250000
8 2700 68 183600 7290000
Total 17200 528 1147800 38500000
Mean of x is
[tex]\bar x = \frac{17200}{8} =2150[/tex]
Mean of y is
[tex]\bar y = \frac{528}{8} =66[/tex]
From the table above
we find [tex]\hat B_1[/tex]
[tex]\hat B_1=\frac{\sum xy- \bar x \sum y}{\sum x^2- n \barx^2} \\\\=\frac{1147800-2150(528)}{38500000-8(2150)^2} \\\\=\frac{1147800-1135200}{38500000-36980000} \\\\=\frac{12600}{1520000} \\\\=0.008289[/tex]
so [tex]\hat b_0[/tex] is
[tex]\hat b_0=\bar y-\bar B_1 x\\\\=66-0.008289(2150)\\\\=66-17.82135\\\\=48.17865[/tex]
The line of regression is
The price x is 1900
[tex]\hat y =\hat B_0+\hat B_1x\\\\=48.17865+0.008289\times1900\\\\=48.17865+15.7491\\\\=63.928[/tex]
The line of regression is 63.928
Answer:
y=48.2+0.00829x;64
Step-by-step explanation:
In a survey of students, 60% were in high school and 40% were in middle school. Of the high school students, 30% had visited a foreign country. If a surveyed student is selected at random, what is the probability that the student is in high school and has visited a foreign country?
Answer:
The probability that the student is in high school and has visited a foreign country is 0.18.
Step-by-step explanation:
We are given that in a survey of students, 60% were in high school and 40% were in middle school.
Of the high school students, 30% had visited a foreign country.
Let the Probability that students were in high school = P(H) = 60%
Probability that students were in middle school = P(M) = 40%
Also, let F = event that students had had visited a foreign country
So, Probability that high school students had visited a foreign country = P(F/H) = 30%
Now, probability that the student is in high school and has visited a foreign country is given by = Probability that students were in high school [tex]\times[/tex] Probability that high school students had visited a foreign country
= P(H) [tex]\times[/tex] P(F/H)
= 0.60 [tex]\times[/tex] 0.30
= 0.18 or 18%
For which x is f(x)?=-3
-7
-4
4
5
Answer:
B.
✔ -4
Step-by-step explanation:
E 2021
New York City is known for it's tourist attractions and high priced real estate. The mean hotel room rate is $202 per night. Assume that the room rates are normally distributed with a standard deviation of $70.What is the probability that a hotel room costs between $210 and $290?
Answer:
[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]
And we can find this probability with this difference
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]
And we can find the difference with the normal standard distirbution or excel:
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]
Step-by-step explanation:
Let X the random variable that represent the hotel room cost of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(202,70)[/tex]
Where [tex]\mu=202[/tex] and [tex]\sigma=70[/tex]
We are interested on this probability
[tex]P(210<X<290)[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using the formula we got:
[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]
And we can find this probability with this difference
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]
And we can find the difference with the normal standard distirbution or excel:
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]
Find the population mean or sample mean as indicated.
Sample: 17, 11, 8, 12, 22
Answer:
mean:12
Step-by-step explanation:
The population mean or sample mean as indicated in the given samples is 14
What is mean?A mean in math is the average of a data set, found by adding all numbers together and then dividing the sum of the numbers by the number of numbers.
Mathematically,
Mean = Sum of the observations/number of observations
Now the given sample is,
17, 11, 8, 12, 22
So, Number of sample = 5
Thus, Mean = Sum of the sample /number of sample
Mean = (17 + 11 + 8 + 12 + 22) / 5
⇒ Mean = 70/5
⇒ Mean = 14
Thus, the population mean or sample mean as indicated in the given samples is 14
To learn more about mean :
https://brainly.com/question/21479395
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Help Me PLEASE!!!
A card is chosen at random from a standard deck of 52 cards, and then it is replaced and another card is chosen. What is the probability that at least one of the cards is a diamond or an ace?
Answer:
P = 0.5207
Step-by-step explanation:
First, we have three options: Just the first card is a diamond or an ace, Just the second card is a diamond or an ace and both cards are diamonds or aces.
Additionally, there are 16 cards that are diamond or aces in a standard deck of 52 cards (13 diamonds and 3 aces that are not diamonds). It means that there are 36 cards that are not diamond or aces (52 - 16 = 36).
So, the probability that just the first card is a diamond or an ace is calculated as:
[tex]P_1=\frac{16}{52}*\frac{36}{52}=0.2130[/tex]
At the same way, the probability that just the second card is a diamond or an ace is:
[tex]P_2=\frac{36}{52}*\frac{16}{52}=0.2130[/tex]
Finally, the probability that both cards are diamonds or aces is:
[tex]P_3=\frac{16}{52}*\frac{16}{52}=0.0947[/tex]
Therefore, the probability that at least one of the cards is a diamond or an ace is:
[tex]P=P_1+P_2+P_3\\P=0.2130+0.2130+0.0947\\P=0.5207[/tex]
2. En la ciudad de Quito, en la temporada fría, se registran temperaturas que van desde los 5 °C hasta los 18 °C. En la temporada cálida, el registro de la temperatura va desde los 4 °C hasta los 30 °C.
a. Representamos estas temperaturas en forma de intervalo y como conjunto.
b. ¿A qué intervalo pertenece la temperatura de la ciudad de Quito?
c. ¿Qué temperaturas son comunes en las temporadas fría y cálida?
d. ¿Qué temperaturas son posibles solo en la temporada fría?
e. ¿Qué temperaturas son posibles solo en la temporada cálida?
Answer:
(See explanation below for further detail/Véase la explicación abajo para mayores detalles)
Step-by-step explanation:
(This exercise is written in Spanish and explanations will be held in such language)
a) Las temperaturas quedan representadas a continuación:
Quito - Temporada Fría
Intervalo
[tex]5 ^{\circ}C \leq t \leq 18^{\circ}C[/tex] (Este intervalo indica si el dato puede pertenecer a la temporada fría)
Conjunto
[tex]C = \{\forall t \in \mathbb {R}| 5 \leq t \leq 18\}[/tex] (Este conjunto acumula todo el registro de las temperaturas de la temporada fría)
Quito - Temporada Cálida
Intervalo
[tex]4 ^{\circ}C \leq t \leq 30^{\circ}C[/tex] (Este intervalo indica si el dato puede pertenecer a la temporada cálida)
Conjunto
[tex]H = \{\forall t \in \mathbb {R}| 4 \leq t \leq 30\}[/tex] (Este conjunto acumula todo el registro de las temperaturas de la temporada cálida)
b) La temperatura de la ciudad de Quito pertenece esencialmente a dos intervalos:
Intervalo de Temporada Fría:
[tex]5 ^{\circ}C \leq t \leq 18^{\circ}C[/tex]
Intervalo de Temporada Cálida:
[tex]4 ^{\circ}C \leq t \leq 30^{\circ}C[/tex]
c) Toda temperatura mayor o igual que 4 °C y menor o igual que 30 °C.
d) Temperaturas mayores o iguales a 5 °C y menores o iguales a 18 °C.
e) Temperaturas mayores o iguales a 4 °C y menores o iguales a 30 °C.
Daniel deposits $300 into an account that earns 16% interest annually. Which equation can be used to model his account balance, y, after x years?
Answer:
[tex]y=300(1+0.16)^x[/tex]
Step-by-step explanation:
This account can be modeled using the compound interest formula.the compound interest formula is expressed as[tex]A= P(1+r )^t[/tex]
Where
A =final amount = y
P=initial principal balance = $300
r=interest rate = 16%= 0.16
t=number of time periods elapsed= x
Hence the equation to model his account balance/ final amount A (y) after time (x) years is
[tex]y=300(1+0.16)^x[/tex]
6(4x - 3) - 30
24x - 18 = 30
24% -18 + 18 = 30 + 18
24x = 48
24x 48
24 24
X = 2
Original Equation
Step 1
Step 2
Step 3
Step 4
Step 5
Which of these is not part of the solution process?
A. Using the associative property
B. Adding 18 to both sides to isolate the variable term
C. Dividing both sides by 24 to isolate the variable
D. Using the distributive property
Answer:
A-using the associative property
Step-by-step explanation:
Plz help me :(
Prove for any value of y the value of expression y^4−(y^2−9)(y^2+9) is divisible by 9.
PROOF:
[tex]y^4-(y^2-9)(y^2+9)\\=y^4 -(y^4-81)\\=y^4-y^4+81\\=81[/tex]
And 81 is ALWAYS divisible by 9
Answer:
Its 9*9
Step-by-step explanation:
The equation equals 81 and 9*9=81.
Have a great day! :)
Please answer this correctly
Answer:
326
Step-by-step explanation:
l x w
7x8
25x6
4x30
326
Determine 6m 9m how much greater the area of the yellow rectangle is than the area of the gree rectangle 2m 5m
Step-by-step explanation:
multiple 6 by 9 then 2 by 5 then subtract them
Answer:
44
Step-by-step explanation:
(6*9) - (2*5)
54 - 10
44
What’s the correct answer for this?
Answer:
D.
Step-by-step explanation:
Since opposite angles of a quadrilateral inscribes in a circle add up to 180°
So,
<P + <N = 180°
2x+2x-12 = 180°
4x = 180+12
4x = 192
Dividing both sides by 4
x = 48
Now
<P = 2(48)
<P = 96
Now
<N = 2(48)-12
<N = 96-12
<N = 84
Watermelon A is 2 kg lighter than watermelon B and it weighs one fifth of the weight of watermelon C. Watermelons A and C together are 3 times as heavy as watermelon B. How heavy is each watermelon?
Answer:
A: 2 kgB: 4 kgC: 10 kgStep-by-step explanation:
We can write some equations to describe the given relationships:
A = B - 2 . . . . . A is 2 kg lighter than B
A = C/5 . . . . . . A is 1/5 the weight of C
A+C = 3B . . . . together, A and C are 3 times the weight of B
__
Let's solve for A.
B = A+2 . . . from the first equation
C = 5A . . . . from the second equation
A +5A = 3(A+2) . . . . substituting for B and C in the third equation
6A = 3A +6
3A = 6
A = 2
__
B = A+2 = 4
C = 5A = 10
Watermelon A weighs 2 kg; B weighs 4 kg; and C weighs 10 kg.
Answer:
2 kg 4 kg 10 kg
I BLESS THE EYES!
Temperature transducers of certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data:
2 1 2 3 1 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
0 4 2 1 3 1 1 3 41 2 3 2 2 8 4 5 1 3 1
5 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 2 3
a. Determine frequencies and relative frequencies for the observed values of x = number of non-conforming transducers in a batch. (Round your relative frequencies to three decimal places.)
b. What proportion of batches in the sample have at most four non-conforming transducers? (Round your answer to three decimal places.)
Answer:
a.
Number: 0, 1, 2, 3, 4, 5, 6, 7, 8
Frequency: 6, 12, 13, 15, 5, 3, 3, 1, 1
b. The proportion of the batches that have at most is 0.864
Step-by-step explanation:
a. The given data are;
2 1 2 3 1 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
0 4 2 1 3 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
5 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 2 3
The frequencies are;
x fx
0 6
1 12
2 13
3 15
4 5
5 3
6 3
7 1
8 1
The relative frequency are;
x Rfx
0 0.102
1 0.203
2 0.220
3 0.254
4 0.085
5 0.051
6 0.051
7 0.017
8 0.017
b. The proportion of the batches that have at most 4 is given as follows;
The number of the batches that have at most 4 = 6 + 12 + 13 + 15 + 5 = 51
Therefore, the proportion of the batches that have at most 4 = 51 / 59 = 0.864.
Tameka can make 32 beads with 4 sheets of paper. Right now she only has 3 sheets of paper. How many beads can Tameka make?
Answer:
24 beads
Step-by-step explanation:
beads in 4 sheet=32
beads in 1 sheet=32/4=8
beads in 3 sheet=8*3=24
Answer:
Assuming that you can make the same number of beads with 1 sheet of paper, Tameka can make 24 beads with 3 sheets of paper.
what is 2043.666666 rounded to 2 decimal places
Answer:
[tex]2043.67[/tex]
Step-by-step explanation:
Hundredths is at 2 decimal places.
The thousandths place is higher than 5, so add 1 to the hundredths place.
Answer:
2043.67
Step-by-step explanation:
If you’ve ever rounded a number, you would know that if it’s 5 or higher, round it up, and if it’s 4 or lower, round it down. In this case, the second decimal place reads ’6’ which is higher that 5, so we round up. The rest of the numbers stay the same
2043.67
22,056 people went to the baseball game on Sunday. Half as many people came on money. How many people were at the baseball game on Sunday and Monday altogether?
Answer:
33084
Step-by-step explanation:
22056 divided by 2 =11028
altogether (on sunday and monday) the total amount would be..
22056+11028=33084
Answer:
33084
Step-by-step explanation:
If 22056 people came to the game on Sunday and Half as many people came on Monday, you do
22056 divided by 2. this is how many people cam on monday
Add this answer to 22056 and this is how many people came on both days.
A gumball machine has 100 red gumballs. If the red gumballs are 25% of the total number of gumballs, how many gumballs are in the gumball machine?
Answer: 400
Step-by-step explanation:
25% is equal to one quarter (1/4). If theres 100 red gumballs then there must be 300 more gumballs in the machine because a quarter of a number is always even.
Write a quadratic function f whose zeros are −6 and −1.
Answer:
y = (x+6) (x+1) or in quadratic form: y = x² + 7x + 6
Step-by-step explanation:
The present value of a perpetuity paying 1 every two years with first payment due immediately is 7.21 at an annual effective rate of i. Another perpetuity paying R every three years with the first payment due at the beginning of year two has the same present value at an annual effective rate of i + 0.01.
Calculate R.
(A) 1.23
(B) 1.56
(C) 1.60
(D) 1.74
(E) 1.94
Answer:
Step-by-step explanation:
image attached (representing first perpetuity on number line)
Present value is 7.21
[tex]7.21=\frac{1}{1-u^2} \\\\1-\frac{1}{7.21} =u^2\\\\\frac{6.21}{7.21} =(1+i)^{-2}\\\\(1+i)^2=\frac{7.21}{6.21} \\\\(i+1)=\sqrt{\frac{7.21}{6.21} }\\\\ i=\sqrt{\frac{7.21}{6.21} } -1\\\\=0.77511297[/tex]
image attached (representing second perpetuity on number line)
we have ,
[tex]7.21=\frac{Ru}{1-u^3}[/tex]
Here,
[tex]V=\frac{1}{1+i}[/tex]i
i = 0.077511297 + 0.01
[tex]\therefore V =\frac{1}{1.087511295} =(1.087511297)^-^1\\\\7.21=\frac{R(1.087511297)^-^1}{1-(1.087511297)^-^3} \\\\7.21=4.132664645R\\\\R=\frac{7.21}{4.132664645} \\\\R= 1.7446370\approx1.74[/tex]
Therefore, value of R is 1.74