Compare the ratios 2/11 and 3/12

The equation πr²Gives the area,A of a circle with a radius,r,. Solve the function for π

zalisa [80]

$\bf \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad \implies \qquad \cfrac{A}{r^2}=\pi$

7 0

3 years ago

Monday What time is 5 3/4 hours after 11:32 pm?

Marina86 [1]

bearing in mind that the an hour has 15 + 15 + 15 + 15 = 60 minutes, so 15 minutes in 1/4 of an hour, thus 45 minutes is 3/4 of an hour.

now, from 11PM, if we add the 5 hours first, we'll be at 4AM, pass midnight of course.

now let's add the minutes, 32 and then 45, that gives us 77 minutes.

so the time will be 4AM plus 77 minutes, since 60 minutes is 1 hr, so 4AM plus 1 hr and 17 minutes, that'd be 5:17AM.

6 0

3 years ago

A survey of 1,562 randomly selected adults showed that 522 of them have heard of a new electronic reader. The accompanying techn

tester [92]

Answer:

a) We want to test the claim that 35% of adults have heard of the new electronic reader, then the system of hypothesis are.:

Null hypothesis: $p=0.35$

Alternative hypothesis: $p \neq 0.35$

And is a two tailed test

b) $z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326$

c) $p_v =2*P(z$

d) Null hypothesis: $p=0.35$

e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Step-by-step explanation:

Information provided

n=1562 represent the random sample selected

X=522 represent the people who have heard of a new electronic reader

$\hat p=\frac{522}{1562}=0.334$ estimated proportion of people who have heard of a new electronic reader

$p_o=0.35$ is the value to verify

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Part a

We want to test the claim that 35% of adults have heard of the new electronic reader, then the system of hypothesis are.:

Null hypothesis: $p=0.35$

Alternative hypothesis: $p \neq 0.35$

And is a two tailed test

Part b

The statistic for this case is given :

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326$

Part c

We can calculate the p value using the laternative hypothesis with the following probability:

$p_v =2*P(z$

Part d

The null hypothesis for this case would be:

Null hypothesis: $p=0.35$

Part e

The best conclusion for this case would be:

Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

5 0

3 years ago

Is this right or wrong? Put these numbers in order from the smallest to the biggest. -1/5, -2/3, 2, 4.

quester [9]

Answer:no the -2/3 should be before -1/5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A survey of 2,254 American adults indicates that 17% of cell phone owners browse the internet exclusively on their phone rather

GrogVix [38]

Answer:

We conclude that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%.

Step-by-step explanation:

We are given that a survey of 2,254 American adults indicates that 17% of cell phone owners browse the internet exclusively on their phone rather than a computer or other device. According to an online article, a report from a mobile research company indicates that 38 percent of Chinese mobile web users only access the internet through their cell phones.

We have to conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%.

Let p = proportion of Americans who only use their cell phones to access the internet

SO, Null Hypothesis, $H_0$ : p = 38% {means that the proportion of Americans who only use their cell phones to access the internet is same as that of Chinese proportion of 38%}

Alternate Hypothesis, $H_a$ : p $\neq$ 38% {means that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of cell phone owners who browse the internet

exclusively on their phone in a survey of 2,254 adults = 17%

n = sample of adults = 2,254

So, test statistics = $\frac{0.17-0.38}{\sqrt{\frac{0.17(1- 0.17)}{2,254} } }$

= -26.542

Since in the question we are not given with the significance level so we assume it to be 5%. So, at 0.05 level of significance, the z table gives critical values between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie in between the critical values of z so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%.

3 0

3 years ago