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2 years ago

Plz help with this I don’t really get it

Mathematics

1 answer:

Greeley [361]2 years ago

6 0

Answer:

Step-by-step explanation:

Associative Property of Multiplication is a(bc) = (ab)c

d is the only one that follows the rule

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I NEED HELP PLEASE THIS IS DUE IN 14 MINUTES. I’LL MARK YOUR ANSWER BRAINLIEST!!!

Lubov Fominskaja [6]

Answer:

1) 12

2) x=9

3) width=26

4) length=16

Step-by-step explanation:

7 0

3 years ago

A. The following events are mutually exclusive: Living in California and watching American Idol. True or False b. The number of

BlackZzzverrR [31]

Answer:

a) False. Because you can live in California AND watch American Idol at the same time

b) True. Because the number of patients is a whole number, like 1, 2 or 3. There is no 1.5 patient

c) False. The actual probability should become closer to the theoretical probablity

d) True

3 0

2 years ago

The Pew Research Center reported that 73% of Americans who own a cell phone also use text messaging. In a recent local survey, 1

AnnZ [28]

Answer:

$z=\frac{0.775 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=1.433$

Now we can find the p value. Since we have a bilateral test the p value would be:

$p_v =2*P(z>1.433)=0.152$

Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

Do Not reject H0

Step-by-step explanation:

Information provided

n=200 represent the sample size slected

X=155 represent the cell phone owners used text messaging

$\hat p=\frac{155}{200}=0.775$ estimated proportion of cell phone owners used text messaging

$p_o=0.73$ is the value to verify

$\alpha=0.1$ represent the significance level

We need to conduct a z test for a proportion

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true proportion of cell phone owners used text messaging is different from 0.73 so then the system of hypothesis are:

Null hypothesis: $p=0.73$

Alternative hypothesis: $p \neq 0.73$

The statistic to check this hypothesis is given by:

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

Replacing the data given we got:

$z=\frac{0.775 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=1.433$

Now we can find the p value. Since we have a bilateral test the p value would be:

$p_v =2*P(z>1.433)=0.152$

Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

Do Not reject H0

6 0

3 years ago

A shipping company claims that 90% of its packages are delivered on time. Jenny noticed that out of the last 10 packages shipped

Gnoma [55]

Answer:

The probability is;

0.0000003645

Step-by-step explanation:

The probability of packages being early p is 90% = 0.9

The probability of packages being late q will be 1-p = 1-0.9 = 0.1

So the probability of 2 out of 10 random late will be subject to Bernoulli approximation of the Binomial theorem

That will be;

P(X = 2) = 10 C 2 0.9^2 0.1^8

= 0.0000003645

4 0

2 years ago

Read 2 more answers

Write an equation parallel to the y-axis through the point (3,−2)

MatroZZZ [7]

do you have a picture of the graph?

8 0

2 years ago