PLZ HELP ASAP The table records the rainfall in two cities during a particular week. Which statements about the data are true?

Can someone please help me with this

Lubov Fominskaja [6]

Answer:

10) C

11) A

Step-by-step explanation:

3 0

3 years ago

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Will spent $3.75 for 3 pounds of granola. What is his unit rate in dollars per pound?

Lady_Fox [76]

Answer:

$1.25 per pound

Step-by-step explanation:

$3.75/3 = 1 pound price

6 0

3 years ago

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A manufacturer knows that their items have a normally distributed lifespan, with a mean of 15 years, and standard deviation of 3

Gala2k [10]

Answer:

The probability that their mean life will be longer than 13 years = .9943 or 99.43%

Step-by-step explanation:

Given -

Mean $(\nu )$ = 15 years

Standard deviation $(\sigma )$ = 3.7 years

Sample size ( n ) =22

Let $\overline{X}$ be the mean life of manufacturing items

the probability that their mean life will be longer than 13 years =

$P(\overline{X}> 13)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{13 - 15 }{\frac{3.7}{\sqrt{22}}})$ $(Z =\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}})$

= $P(Z> -2.53)$

= $1 - P(Z < -2.53)$ Using Z table

= 1 - .0057

= .9943

7 0

3 years ago

A music store sold 10 to the third power CDs and 10 to the second power cd players if each CD cost $12 in each CD player cost 35

Mamont248 [21]

Answer:

$15500

Step-by-step explanation:

10^3 x 12 = 12,000

10^2 x 35 = 3,500

12,000 + 3,500 = 15,500

6 0

3 years ago

A study by the Pew Research Center1 reports that in 2010, for the first time, more adults aged 18 to 29 got their news from the

sergejj [24]

Answer:

Null hypothesis: $p \leq 0.65$

Alternative hypothesis: $p > 0.65$

$z=\frac{0.66 -0.65}{\sqrt{\frac{0.65(1-0.65)}{1500}}}=0.812$

$p_v =2*P(z>0.812)=0.208$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion it's not significantly higher from 0.65.

We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the adults that said television was one of their main sources of news.

$\hat p=0.66$ estimated proportion of adults that said television was one of their main sources of news.

$p_o=0.65$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 0.65:

Null hypothesis: $p \leq 0.65$

Alternative hypothesis: $p > 0.65$

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.66 -0.65}{\sqrt{\frac{0.65(1-0.65)}{1500}}}=0.812$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>0.812)=0.208$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion it's not significantly higher from 0.65.

Do we find evidence that more than 65% of all US adults use television as one of their main sources for news?

We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.

6 0

3 years ago