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2 years ago

13

in a classroom the number of girls is 3 and 2/3 times the number of boys. there are 16 more girls than boys. find the total numb

er of students in the classroom. Explain your answer.

1 answer:

nalin [4]2 years ago

5 0

I think it is 19 boys in the classroom

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What is 2 15/16 Ian’s a decimal?

Arada [10]

Answer:

2.9375

Step-by-step explanation:

15 divided by 16 equals .9375 so 2 15/16 is 2.9375

5 0

2 years ago

Let f(x) = 1/x . Find the number b such that the average rate of change of f on the interval [2, b] is − 1/8

vekshin1

Answer:

b=4

Step-by-step explanation:

So, we have the function $f(x)=1/x$ . We need to find b such that the average rate of change or the slope is -1/8 between the intervel [2, b]. First, let's find f(2).

f(2) = 1/(2) = 1/2

So, we have the point (2, 1/2)

At point b, f(b) = 1/b.

Let's plug this into the slope formula:

$\frac{y_2-y_1}{x_2-x_1}=\frac{.5-\frac{1}{b} }{2-b} =-1/8$

Now, we just need to solve for b. First, let's multiply both the numerator and denominator by b (to get rid of the annoying fraction in the numerator).

$\frac{.5b-1}{2b-b^2} =\frac{-1}{8}$

Now, cross multiply.

$4b-8=b^2-2b$

$b^2-6b+8=0$

Solve for b. Factor using the numbers -4 and -2.

$=(b-4)(b-2)=0$

Thus, b=4 or b=2.

However, b=2 is not a possible solution since the interval [2,2] means nothing. Thus, b=4.

4 0

2 years ago

Given: △RST ~ △VWX

tatiyna

Answer:

Given

Step-by-step explanation:

Given that: △RST ~ △VWX, TU is the altitude of △RST, and XY is the altitude of △VWX.

Comparing △RST and △VWX;

TU ~ XY (given altitudes of the triangles)

<TUS = <XYW (all right angles are congruent)

<UTS ≅ <YXW (angle property of similar triangles)

Thus;

ΔTUS ≅ ΔXYW (congruent property of similar triangles)

<UTS + <TUS + < UST = <YXW + <XTW + <XWY = $180^{0}$ (sum of angles in a triangle)

Therefore by Angle-Angle-Side (AAS), △RST ~ △VWX

So that:

$\frac{TU}{XY}$ = $\frac{US}{YW}$ (corresponding side length proportion)

4 0

2 years ago

I need helpppp !!! :((

Andrew [12]

Im not 100% sure but I do think it is Ratio (red to blue) of the areas

Step-by-step explanation:

3 0

3 years ago

Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b

sesenic [268]

Answer:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

-The sample is too small to make judgments about skewness or symmetry.

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

$p_v =2*P(t_{(14)}$

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

Step-by-step explanation:

First we need to find the difference defined as:

(Operator 1 minus Operator 2)

d1=1.326-1.323=0.003 d2=1.337-1.322=0.015

d3=1.079-1.073=0.006 d4=1.229-1.233=-0.004

d5=0.936-0.934=0.002 d6=1.009-1.019=-0.01

d7=1.179-1.184=-0.005 d8=1.289-1.304=-0.015

Now we can calculate the mean of differences given by:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

And for the sample deviation we can use the following formula:

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

Describe the distribution of these differences using words. (which one is correct)

We can plot the distribution of the differences with the folowing code in R

differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)

hist(differences)

And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.

-The sample is too small to make judgments about skewness or symmetry.

Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)

$\bar X_{1}=1.173$ represent the mean for the operator 1

$\bar X_{2}=1.174$ represent the mean for the operator 2

$s_{1}=0.1506$ represent the sample standard deviation for the operator 1

$s_{2}=0.1495$ represent the sample standard deviation for the operator 2

$n_{1}=8$ sample size for the operator 1

$n_{2}=8$ sample size for the operator 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) like this:

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(14)}$

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

8 0

3 years ago