What is the LCM of 56 and 504?

What is the mean and median of 73 45 88 90 90 81 83

notsponge [240]

Answer:

Mean = 78.5

Median = 83

3 0

4 years ago

144 divided by 8 long division please help.

romanna [79]

Well 144 divided by 8

7 0

3 years ago

Read 2 more answers

A paper reported that in a representative sample of 282 American teens age 16 to 17, there were 71 who indicated that they had s

elena-s [515]

Answer:

$z=\frac{0.252 -0.25}{\sqrt{\frac{0.25(1-0.25)}{282}}}=0.0776$

$p_v =P(z>0.0776)=0.4691$

Since the p value is higher than the significance level so then we can conclude that we have enough evidence to FAIL to reject the null hypothesis and there is no evidence in order to say that the true proportion of teens more than a quarter of Americans age 16 to 17 have sent a text message while driving is higher than 0.25

Step-by-step explanation:

Information provided

n=282 represent the sample selected

X=71 represent the teens indicated that they had sent a text message while driving

$\hat p=\frac{71}{282}=0.252$ estimated proportion of teens indicated that they had sent a text message while driving

$p_o=0.25$ is the value that we want to test

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to check if more than a quarter of Americans age 16 to 17 have sent a text message while driving, and the hypothesis are:

Null hypothesis: $p\leq 0.7$

Alternative hypothesis: $p > 0.25$

The statistic is given by:

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

Replacing we got:

$z=\frac{0.252 -0.25}{\sqrt{\frac{0.25(1-0.25)}{282}}}=0.0776$

Decision

The p value for this case is:

$p_v =P(z>0.0776)=0.4691$

Since the p value is higher than the significance level so then we can conclude that we have enough evidence to FAIL to reject the null hypothesis and there is no evidence in order to say that the true proportion of teens more than a quarter of Americans age 16 to 17 have sent a text message while driving is higher than 0.25

8 0

4 years ago

A single card is drawn from a standard 52-card deck. Let D be the event that the card drawn is a black card, and let F be the

Svetllana [295]

Answer:

$\dfrac{25}{26}$

Step-by-step explanation:

Let

D be the event that the card drawn is a black card
F be the event that the card drawn is a 10 card
F' be the event that the card drawn is not a 10 card

The probability

$Pr(D\cup F')=Pr(D)+Pr(F')-Pr(D\cap F')\\$

Note that

$Pr(D)=\dfrac{26}{52}=\dfrac{1}{2}\\ \\Pr(F)=\dfrac{4}{52}=\dfrac{1}{13}\\ \\Pr(F')=1-\dfrac{1}{13}=\dfrac{12}{13}\\ \\Pr(D\cap F')=\dfrac{24}{52}=\dfrac{6}{13}$

So,

$Pr(D\cup F')=\dfrac{1}{2}+\dfrac{12}{13}-\dfrac{6}{13}=\dfrac{25}{26}$

8 0

3 years ago

Find the exact value of the trigonometric function of theta from the given information. cosine thetaequalsnegative StartFraction

Iteru [2.4K]

Is it a multiple choise?

7 0

4 years ago