The radius of the large sphere is double the radius of the
small sphere.
How many times is the volume of the large sphere than the
small sphere?
02
O4
6
O 6
O 8
Answers
Answer:
8
Step-by-step explanation:
just do the comparation
Vb : Vs
b for big and s for small
4/3 π rb³ : 4/3 π rs³ (since there are 4/3 and π on both side, we can eliminate them so)
Vb : Vs = rb³ : rs³
Vb : Vs = (2rs)³ : rs³
Vb : Vs = 8rs³ : rs³ (delete the r³ on both side)
Vb : Vs = 8 : 1
so Vb is 8 times larger in volume than the small one
Related Questions
The drama club is selling candles for a fundraiser. They spend $100 on the candles and sell them for $4.50 each. How many candles must they sell to make more than $125 profit?
Let x represent the number of candles sold. Which inequality can you use to find x?
Answers
So I try to help
Step-by-step explanation:
I don't no sorrry
Answer:
the first one!!
Step-by-step explanation:
Write down five numbers with a mode of 6.
Answers
Answer:
6 4 5 6 7
Step-by-step explanation:
The mode would be 5. The first 2 letters of MODE are MO, which can be an abbreviation for MOST OFTEN.
Given that d is the midpoint of line segment ab and k is the midpoint of line segment bc, which statement must be true? (May give brainliest)
Answers
Answer:
B is the midpoint of line segment AC
What are the next two numbers in the pattern of numbers 45,15,44,17,40,20,31,25
Answers
Answer:
14, 32
Step-by-step explanation:
45,15,44,17,40,20,31,25
this is combination of 2 series:
45-44-40-31- ?15-17-20-25-?In the first series we can see the pattern as:
-1, -4, -9 = -1², -2², -3² so next difference must be -4², which is 31- 16= 14In the second series we can see the pattern as:
2, 3, 5 prime numbers, so next difference must be 7, which is 25+7=32The series will continue as:
45, 15, 44, 17, 40, 20, 31, 25, 14, 32Answer:
14, 32
Step-by-step explanation:
lol :D
Find the probability that in 200 tosses of a fair die, we will obtain at most 30 fives
Answers
Answer:
0.2946
Step-by-step explanation:
Number of tosses, n = 200
P(obtaining a 5), p = 1/6
q = 1 - p = 5/6
Normal approximation for binomial distribution
P(X < A) = P(Z < (A - mean)/standard deviation)
Mean = np
= 200 x 1/6
= 33.33
Standard deviation = √npq
= √(200(1/6)(5/6) )
= 5.27
P(at most 30 fives) = P(X ≤ 30)
= P(Z < (30.5 - 33.33)/5.27) (continuity correction of 0.5 is added to 30)
= P(Z < -0.54)
= 0.2946
We wish to find the probability that a child from this population who has inadequate calcium intake is 11 to 13 years old. In other words, if you know that a child has inadequate calcium intake, what is the probability that the child is between 11 and 13 years old
Answers
Answer:
Step-by-step explanation:
Look at the population statistics. Let's say it contains:
- data on the age groups available in the population
- data on the probability that a child in the population has inadequate calcium intake OR data that a child in the population does not have the deficiency. If you're given one of these, the other can be gotten by subtracting the probability value given from 1.
So let's say there are children from ages 5 to 15 in this population and the probability that a child in this population has the deficiency is 0.23 (not all the children in this population of 5-15 year olds may have the deficiency) while the probability that a child in this population does not have the deficiency is [1-0.23] = 0.77
So if you pick a child randomly from the population and he has this deficiency, what is the probability that he or she is between 11 and 13 years old?
From ages 5-15, ages 11, 12 and 13 are 3 ages. The total number of ages is 11 ages.
3÷11 = 0.2727
This is the probability that a child picked or selected at random from the population is 11, 12, or 13 years old.
0.2727 × 0.23 = 0.0627
This is the probability that a child picked at random is BOTH within the age bracket 11 to 13 AND has the deficiency!
Apply this.
The mean percent of childhood asthma prevalence in 43 cities is 2.32%. A random sample of 32 of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8%? Interpret this probability. Assume that sigmaequals1.24%. The probability is nothing.
Answers
Answer:
[tex] P(\bar X>2.8)[/tex]
We can use the z score formula given by:
[tex] z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z=\frac{2.8 -2.32}{\frac{1.24}{\sqrt{32}}}=2.190 [/tex]
And using the normal standard distribution and the complement rule we got:
[tex] P(z>2.190 )= 1-P(z<2.190) = 1-0.986=0.014[/tex]
Step-by-step explanation:
For this case w eknow the following parameters:
[tex] \mu = 2.32[/tex] represent the mean
[tex]\sigma =1.24[/tex] represent the deviation
n= 32 represent the sample sze selected
We want to find the following probability:
[tex] P(\bar X>2.8)[/tex]
We can use the z score formula given by:
[tex] z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z=\frac{2.8 -2.32}{\frac{1.24}{\sqrt{32}}}=2.190 [/tex]
And using the normal standard distribution and the complement rule we got:
[tex] P(z>2.190 )= 1-P(z<2.190) = 1-0.986=0.014[/tex]
Answer:
0.55% probability that the mean childhood asthma prevalence for the sample is greater than 2.8%. This means that a sample having an asthma prevalence of greater than 2.8% is unusual event, that is, unlikely.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
If X is more than two standard deviations from the mean, it is considered an unusual outcome.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 2.32, \sigma = 1.24, n = 43, s = \frac{1.24}{\sqrt{43}} = 0.189[/tex]
What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8%?
This is 1 subtracted by the pvalue of Z when X = 2.8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.8 - 2.32}{0.189}[/tex]
[tex]Z = 2.54[/tex]
[tex]Z = 2.54[/tex] has a pvalue of 0.9945
1 - 0.9945 = 0.0055
0.55% probability that the mean childhood asthma prevalence for the sample is greater than 2.8%. This means that a sample having an asthma prevalence of greater than 2.8% is unusual event, that is, unlikely.
Suppose point (4, −9) is translated according to the rule (, ) → ( + 3, − 2). What are the coordinates of ′? Explain.
Please help
Answers
Answer:
(7, -11)
Step-by-step explanation:
If the point is shifted 3 to the right and 2 down, you just have to add 3 to the x-coordinate and subtract 2 from the y-coordinate. 4+ 3 = 7 and -9 - 2 is -11. So, the new point will be (7, -11).
Answer:
(7, -11)
Step-by-step explanation:
The point is translated three units to the right, and 2 units down.
[tex](4,-9)=>(4+3,-9-2)=>(7,-11)[/tex]
Point " ' " should be (7,-11)
Erythropoietin (EPO) is a banned drug used by athletes to increase the oxygen-carrying capacity of their blood. New tests for EPO were first introduced prior to the 2000 Olympic Games held in Sydney, Australia. Chance (Spring 2004) reported that of a sample of 830 world-class athletes, 159 did not compete in the 1999 World Championships (a year prior to the new EPO test). Similarly, 133 of 82:5 potential athletes did not compete in the 2000 Olympic Games. Was the new test effective in deterring an athlete's participation in the 2000 Olympics? If so, then the proportion of nonparticipating athletes in 2000 will be greater than the proportion of nonparticipating athletes in 1999, Use a 98% confidence interval to compare the two proportions and make the proper conclusion.
Answers
Answer:
Step-by-step
The null and the alternative hypothesis can be define as follows,
Null Hypothesis; There is no significance difference between the proportions of non participating athletes in 1999 and 2000
[tex]H_0:(p_1-p_2)\neq 0[/tex]
Alternative Hypothesis: The proportion of non participating athletes in 2000 will be more than the proportion of non participating athletes in 1999
[tex]H_1:(p_1-p_2)<0[/tex]
The proportion of nonparticipating athletes in 1999 is given by
[tex]\hat p_1 = \frac{x_1}{n_1} \\\\=\frac{159}{830} =0.1916[/tex]
The proportion of nonparticipating athletes in 2000 is given by
[tex]\hat p_2 =\frac{x_2}{n_1} \\\\=\frac{133}{825} =0.1612[/tex]
The pooled proportion can be calculated using the following formula
[tex]\hat p = \frac{x_1+x_2}{n_1+n_2} \\\\=\frac{159+133}{830+825} =0.1764[/tex]
under the null hypothesis, the test statistics can be calculated as follows
[tex]Z=\frac{\hat p_1 - \hat p_2}{\sqrt{\hat p \hat q(\frac{1}{n_1}+\frac{1}{n_2} ) } }[/tex]
[tex]=\frac{0.1916-0.1612}{\sqrt{(0.1764)(0.8236)(\frac{1}{830} +\frac{1}{825} )} } \\\\=1.6257[/tex]
Determine the P-value using the following formula
P-value = Normdist(1.6257)
=0.947993
Here, it can be observed that the P-value is greater than the level of the significance,
Hence, the null hypothesis fails to be rejected
Therefore it can be concluded that there is insufficient evidence to support that the proportions of non participating athletes in 2000 will be more than the proportions of non participating athletes in 1999
Find the area of a circle with radius, r = 9cm.
Give your answer in terms of π .
Answers
Answer:
[tex]81\pi[/tex]
Step-by-step explanation:
[tex]Area = \pi * r^{2} \\\pi *9^{2} =81\pi[/tex]
Answer:
81 π
Step-by-step explanation:
formula is radius squared times pi or π so the answer would be 9x9=81 and you said to leave in terms of π so the answer is 81 π.
Can you help me with this one don’t get it
Answers
Do you need an explication?
The total degrees of angles is (n-2) *180
And since there are 5 sides it’s 3 *180 which gives u 540. Then you subtract all the given angles or find the remaining angle
Please answer this correctly
Answers
Answer:
Height of this missing bar would be 1
Step-by-step explanation:
Since there is 1 and only 1 quantity between 80-99.
Answer:
1
There is 1 number that is between 80 and 99 which is 99 so there should be 1 bar.
Step-by-step explanation:
Round 90.2844097979 to 3 decimals
Answers
Answer:
only allow 3 decimals
90.284 is the answer we removed all others except for 3
1.solve for x 3(10 - 2x)=18
Answers
Answer:
[tex]\boxed{\ x=2\ }[/tex]
Step-by-step explanation:
3(10-2x)=18
<=>
10-2x=18/3=6
<=>
2x=10-6=4
<=>
x= 4/2=2
Lesson 9: Problem Solving When the Percent Changes
Exit Ticket
Tamia and Laniece were selling magazines for a charity. In the
first week, Tamia sold 30% more than Laniece. In the second
week, Tamia sold 12 magazines, but Laniece did not sell any. If
Tamia sold 50% more than Laniece by the end of the second
week, how many magazines did Laniece sell? Choose any
model to solve the problem. Show your work to justify your
answer.
Answers
Answer:
Laniece had 60 magazines
Step-by-step explanation:
Given: In the first week, Tamia sold 30% more than Laniece. In the second week, Tamia sold 12 magazines, but Laniece did not sell any. Tamia sold 50% more than Laniece by the end of the second week
To find: Number of magazines sold by Laniece
Solution:
Let number of magazines sold by Laniece in the first week be x.
Number of magazines sold by Tamia in the first week = [tex]x+\frac{30}{100} x=\frac{130x}{100} =\frac{13x}{10}[/tex]
Number of magazines sold by Tamia in the second week = 12
Total number of magazines sold by Tamia at the end of the second week = [tex]\frac{13x}{10}+12[/tex]
Total number of magazines sold by Laniece at the end of the second week = x
According to question,
[tex]\frac{13x}{10}+12=x+\frac{50x}{100}=x+\frac{x}{2}\\\frac{13x}{10}+12=\frac{3x}{2}\\\frac{3x}{2}-\frac{13x}{10} =12\\\frac{15x-13x}{10}=12\\\frac{2x}{10}=12\\\frac{x}{5}=12\\x=60[/tex]
The HCF of two numbers is 11, and their L.C.M is 368. If one number is 64, then the other number is
Answers
Answer:
63.25 not an integer
Step-by-step explanation:
HCF(a,b)*LCM(a,b)=ab
11*368=64*x
x=11*368/64
x=63.25 not an integer, one of the given numbers must be incorrect
but you may use this method to find it yourself
Find the range of the set of numbers shown by the box-and-whisker plot.
6
6
7
8
9
10
11
12
13
14
16
16
17
18
19
20
21
22
Answers
Answer:
16
Step-by-step explanation:
The range is the difference between the highest and lowest number of a data set. 22-6=16
A circle with circumference 6 has an arc with a 20 degrees central angle. What is the length of the arc?
Answers
Answer:
[tex]\frac{1}{3}[/tex]
Step-by-step explanation:
So first calculate what fraction of the circumference the arc is.
[tex]\frac{20}{360}=\frac{1}{18}[/tex]
Now the circumference is 6, so one eighteenth of that is [tex]\frac{1}{3}[/tex]
If the mean of 5 positive integers is 15, what is the maximum possible difference between the largest and the smallest of these 5 numbers?
Answers
Answer:
64
Step-by-step explanation:
If the mean is 15, the sum of 5 numbers is:
5*15 = 75Minimum value for the first four numbers would be:
1, 2, 3, 4Then the fifth number is:
75 - (1+2+3+4) = 75 - 10 = 65So the maximum difference is:
65 - 1 = 64The length of a rectangle is 4 inches longer than the width. If the area is 390 square inches, find the rectangle's dimensions. Round your answers to the nearest tenth of an inch.
Answers
Answer:
17.8 in x 21.8 in
Step-by-step explanation:
Given w=width and l=length
w*l=390
l=w+4, therefore w*(w+4)=390
w^2+4w=390
w^2+4w-390=0
Quadratic equation, solve as such
w=-21.8 or 17.8
Solution can't be negative so w=17.8 in
l=w+4 so l=21.8
60 is what percent of 400
Answers
Answer:
15%
Step-by-step explanation:
Is means equals and of means multiply
60 = P * 400
Divide each side by 400
60/400 = P
.15 = P
Change to percent form
15% is the percent
Answer:
the answer to the question you've asked is 15
Playbill magazine reported that the mean annual household income of its readers is $119,155 (Playbill, January 2006). Assume this estimate of the mean annual household in- come is based on a sample of 80 households, and based on past studies, the population standard deviation is known to be a = $30,000. a. Develop a 90% confidence interval estimate of the population mean. b. Develop a 95% confidence interval estimate of the population mean. c. Develop a 99% confidence interval estimate of the population mean. d. Discuss what happens to the width of the confidence interval as the confidence level is increased. Does this result seem reasonable? Explain.
Answers
Answer:
a) CI = (113,637.5 , 124,672.5)
b) CI = (112,581 , 125,729)
c) CI = (110,501.4 , 127,808.6)
Step-by-step explanation:
You have the following information:
[tex]\overline{x}[/tex]: mean annual household income = 119,155
σ: standard deviation = 30,000
n: sample = 80
The interval of confidence is given by the following expression:
[tex]\overline{x}\pm Z_{\alpha/s}(\frac{\sigma}{\sqrt{n}})[/tex]
Z_α/2: distribution density factor
where α and Z_α/2 are given by the range of the confidence interval.
a) For a 90% confidence interval you have:
α = 1 - 0.9 = 0.1
Z_0.1/2 = Z_0.05 = 1.645 (found in a table of normal distribution)
You replace in the equation (1) to obtain the confidence interval:
[tex]119,155\pm (1.645)(\frac{30,000}{\sqrt{80}})\\\\=119,155\pm5,517.5[/tex]
Then, the confidence interval is (119,155 + 5,517.5 , 119,155 - 5,517.5 )
= (113,637.5 , 124,672.5)
b) For a 95% confidence interval you have:
α = 1 - 0.95 = 0.05
Z_0.05/2 = Z_0.025 = 1.96
[tex]119,155\pm (1.96)(\frac{30,000}{\sqrt{80}})\\\\=119,155\pm 6,574.0[/tex]
The confidence interval is (112,581 , 125,729)
c) For a 99% confidence interval:
α = 1 - 0.99 = 0.01
Z_0.01/2 = Z_0.005 = 2.58
[tex]119,155\pm (2.58)(\frac{30,000}{\sqrt{80}})\\\\=119,155\pm 8,653.6[/tex]
The confidence interval is (110,501.4 , 127,808.6)
d) When the confidence level increases the width of the confidence increases too. This can be noticed in the normal distribution, when the confidence level is higher, the area of the tails is reduced, and so, the confidence interval is higher.
A study of the effect of smoking on sleep patterns is conducted. The measure observed is the time, in minutes, that it takes to fall asleep. These data are obtained:
Smokers: 69.3 56.0 22.1 47.6 53.2 48.1 52.7 34.4 60.2 43.8 23.2 13.8
Non-Smokers: 28.6 25.1 26.4 34.9 28.8 28.4 38.5 30.2 30.6 31.8 41.6 21.1 36.0 37.9 13.9
Which group having greater value of relative dispersion and why?
Answers
Answer:
The group that has greater value of relative dispersion is the smokers group, as the coefficient of variationof their data is bigger than the coefficient of variation of the non-smokers group data.
CV smokers: 0.387
CV non-smokers: 0.234
Step-by-step explanation:
We will calculate the relative dispersion of each data set with its coefficient of variation (ratio of the standard deviation to the arithmetic mean).
Then, first we calculate the mean and standard deviation for the smokers data:
Mean: 43.7
Standard deviation: 286.5
[tex]M_s=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M_s=\dfrac{1}{12}(69.3+56+22.1+47.6+53.2+. . .+13.8)\\\\\\M_s=\dfrac{524.4}{12}\\\\\\M_s=43.7\\\\\\s_s=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M_s)^2\\\\\\s_s=\dfrac{1}{11}((69.3-43.7)^2+. . . +(13.8-43.7)^2)\\\\\\s_s=\dfrac{3152}{11}\\\\\\s_s=286.5\\\\\\[/tex]
The mean and standard deviation for the non-smokers is:
Mean: 30.3
Standard deviation: 50.9
[tex]M_n=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M_n=\dfrac{1}{15}(28.6+25.1+26.4+34.9+28.8+. . .+13.9)\\\\\\M_n=\dfrac{453.8}{15}\\\\\\M_n=30.3\\\\\\s_n=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M_n)^2\\\\\\s_n=\dfrac{1}{14}((28.6-30.3)^2+. . . +(13.9-30.3)^2)\\\\\\s_n=\dfrac{713.3}{14}\\\\\\s_n=50.9\\\\\\[/tex]
Now, we can calculate the coefficient of variation:
CV smokers:
[tex]CV_s=\dfrac{s_s}{M_s}=\dfrac{16.9}{43.7}=0.387[/tex]
CV non-smokers:
[tex]CV_n=\dfrac{s_n}{M_n}=\dfrac{7.1}{30.3}=0.234[/tex]
A survey showed that 82% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 15 adults are randomly selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight correction?
Answers
Answer:
[tex] P(X \leq 1)= P(X=0) +P(X=1) [/tex]
And using the probability mass function we can find the individual probabiities
[tex]P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}[/tex]
[tex]P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}[/tex]
And replacing we got:
[tex] P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}[/tex]
And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight correction in a sample of 15 since the probability obtained is very near to 0
Step-by-step explanation:
Let X the random variable of interest "number of adults who need correction", on this case we now that:
[tex]X \sim Binom(n=15, p=0.82)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
We want to find this probability:
[tex] P(X \leq 1)= P(X=0) +P(X=1) [/tex]
And using the probability mass function we can find the individual probabiities
[tex]P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}[/tex]
[tex]P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}[/tex]
And replacing we got:
[tex] P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}[/tex]
And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight correction in a sample of 15 since the probability obtained is very near to 0
Which graph represents this equation y-4= -3(x+5)
Answers
Answer:
Graph B
Step-by-step explanation:
Simplify.
y - 4 = -3x - 15 Distribute
y = -3x - 11 Add 4 on both sides
The y-intercept should be negative, and option B has a negative y-intercept.
The graph of the given function will be represented by graph B so the correct answer is option B.
What is a graph?A graph is the representation of the data on the vertical and horizontal coordinates so we can see the trend of the data.
The graph of the function is attached with the answer below.
Simplify.
y - 4 = -3x - 15 Distribute
y = -3x - 11 Add 4 on both sides
The y-intercept should be negative, and option B has a negative y-intercept.
Therefore the graph of the given function will be represented by graph B so the correct answer is option B.
To know more about graphs follow
https://brainly.com/question/4025726
SPJ5
Saved
250 mg
sing value in
50 mg
10 ml
X
Choice
Answers
What is the question and then i can help!
If f(x)=7+4c and g(x) = 1/2x what is the value of (f/g)(5)
Answers
Answer: 270
Step-by-step explanation:
The notation [tex](\frac{f}{g})(5)[/tex] means to divide [tex]\frac{f(5)}{g(5)}[/tex]. Now that we know we have to divide, we can plug them into this equation.
[tex]\frac{7+4(5)}{\frac{1}{2(5)} }=\frac{27}{\frac{1}{10} }[/tex]. We know that dividing by a fraction means to multiply by its reciprocal, so we'll do that.
[tex]27*10=270[/tex]
Determine whether the stated causal connection is valid. If the causal connection appears to be valid, provide an explanation. Test grades are affected by the amount of time and effort spent studying and preparing for the test. Choose the correct answer below
a. The causal connection is valid. Students who spend more time and effort studying will be able to memorize more information, so their test grades will be higher.
b. The causal connection is valid. Students who spend more time and effort studying tend to be smarter, so their test grades are higher.
c. The causal connection is valid. When students spend more time and effort studying for a test, their test grades tend to be higher.
d. The causal connection is not valid.
Answers
Answer:
A. The causal connection is valid. Students who spend more time and effort studying will be able to memorize more information, so their test grades will be higher.
Step-by-step Explanation:
The causal connection between the test grades of students and the amount of time and effort spent the students spend in studying and preparing for the test appears to be valid. This is valid because students who spend more time and effort studying would most likely be able to memorize more information of which they are most likely to come by in the test they take. Invariably, they'd be able to easily recall what they've memorize and give the right answers to the questions they are asked in the test, and this definitely will earn them higher test grades.
I will mark you as BRANLIEST and I will give you 55 points if you answer correctly.
~Solve each system of equations using elimination. SHOW YOUR WORK
1) -4x+3y=-5
4x-5y=3
2) 2x+3y=36
10x-6y=12
Answers
Answer:
1. x = 2 y = 1
2. x = 17 y = 29/3
Step-by-step explanation:
1. Use elimination
-4x+3y=-5
4x-5y=3
x's cancel, so add them
-2y = -2
y = 1
substitute
4x -5(1) = 3
4x - 5 = 3
4x = 8
x = 2
2. Use elimination
2x+3y=36
10x-6y=12
Multiply top equasion by 5
10x+30y = 360
10x-6y = 12
x's cancel so subract
36y = 348
y = 29/3
Substitute
2x+3(2/3)=36
2x+2 = 36
2x = 34
x = 17
Please answer this correctly
Answers
Answer:
12 2/5 hours
Step-by-step explanation:
[tex]1+1+1\frac{1}{5} +1\frac{1}{5} +1\frac{1}{5} +1\frac{3}{5} +1\frac{3}{5} +1\frac{4}{5} +1\frac{4}{5} =\\\\2+3\frac{3}{5} +3\frac{1}{5} +3\frac{3}{5} =\\\\11\frac{7}{5} =\\\\12\frac{2}{5}[/tex]
12 2/5 hours have been logged in all.
