All categories

3 years ago

These are the means and standard deviations for samples of building heights

Mathematics

1 answer:

Ulleksa [173]3 years ago

4 0

Answer: A & C

A- City A’s heights are more spread out than city B’s heights.

C- City A has a lower average height than city B.

You might be interested in

C=2(3.14)r; c=14(3.14) solve for r

tekilochka [14]

Answer:

c=2(3.14)r

given c=14(3.14), then r

make r the subject formula and substitute given c=14(3.14)

r=c/2(3.14)

r=14(3.14)/2(3.14)

r=7

Step-by-step explanation:

7 0

3 years ago

A group of students are standing in rows. There are as many students in a row as there are rows. If there are 64 students in the

Zanzabum

Answer:

B..im pretty sure

Step-by-step explanation:

First it cant be D because 12 x __ = 64. There is no whole number multiplied by 12 that equals 64. So cross that out. then it's not B because 6 x ___ = 64. So also cross that out. the only thing u have left is 4 and 8. Then u can cross out A because the group of students has to be the same number of rows. Which is 8 x 8 aka B. The answer is B

4 0

2 years ago

Read 2 more answers

Decrease<br> M720.00 by 25%

netineya [11]

Answer:

540

Step-by-step explanation:

720 (100-25) to find the decimal to use (below 1.00 is decreasing)

720 (.75) = 540

8 0

3 years ago

Multiply the additive inverse of (-7/8) by the reciprocal of (5 1/2)<br><br>PLEASE HELP!

EleoNora [17]

Answer:

$\bold {$\dfrac{7}{44}$}$

Step-by-step explanation:

<h3> Reciprocal:</h3>

$\sf \text{The reciprocal of any number $\dfrac{a}{b}$ is given by $\dfrac{b}{a}$}\\\\5\dfrac{1}{2}=\dfrac{11}{2}\\\\\text{Reciprocal of $\dfrac{11}{2}$ = $\dfrac{2}{11}$}$

<h3>Additive inverse:</h3>

$\text{Additive inverse of $\dfrac{-7}{8}$=$\dfrac{7}{8}$}$

Now multiply,

$\sf \dfrac{7}{8}*\dfrac{2}{11}=\dfrac{7}{4}*\dfrac{1}{11}\\$

$=\dfrac{7}{44}$

7 0

1 year ago

The Genetics & IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceivi

Reika [66]

Answer: The test statistic needed to test this claim= 10.92

Step-by-step explanation:

We know that the probability of giving birth to a boy : p= 0.5

i..e The population proportion of giving birth to a boy = 0.5

As per given , we have

Null hypothesis : $H_0: p\leq0.5$

Alternative hypothesis : $H_a: p>0.5$

Since $H_a$ is right-tailed , so the hypothesis test is a right-tailed z-test.

Also, it is given that , the sample size : n= 291

Sample proportion: $\hat{p}=\dfrac{239}{291}\approx0.82$

Test statistic : $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$ , where n is sample size , $\hat{p}$ is sample proportion and p is the population proportion.

$\Rightarrow\ z=\dfrac{0.82-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{291}}}\approx10.92$

i.e. the test statistic needed to test this claim= 10.92

Critical value ( one-tailed) for 0.01 significance level = $z_{0.01}=2.326$

Decision : Since Test statistic value (10.92)> Critical value (2.326), so we reject the null hypothesis .

[When test statistic value is greater than the critical value , then we reject the null hypothesis.]

Thus , we concluded that we have enough evidence at 0.01 significance level to support the claim that the YSORT method is effective in increasing the likelihood that a baby will be a boy.

7 0

3 years ago