WHAT IS 4.9% AS A DECIMAL AND A FRACTION

Solve the equation for x. 10x= 120

Dominik [7]

Answer:

x = 12

Step-by-step explanation:

10x = 120 original equation

10x/10 = 120/10 divide both sides by 10

x = 12 simplify

Check work:

10(12) = 120

120 = 120 (true)

6 0

3 years ago

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True or false??????

Katyanochek1 [597]

Answer:

true

Step-by-step explanation:

$\sqrt{25} = ±5$

6 0

3 years ago

1. A senator is thinking about running for president. But she will only do so if more than 60% of the voters view her favorably.

OLga [1]

The senator is faced with the making a decision to run for President, based

on the conclusion she can make from the data she gathers.

The statistical method that would be best for her to use is inferential

statistic method such as finding the percentage of supporters she has to a

given confidence level.

Reason:

The two main types of statistical methods are;

Inferential statistics methods; Theses are methods used to reach a

conclusion that extends past the given or available data.

Descriptive statistics methods; Theses are statistical method used as a

summary of a data set.

Given that the senator is trying to draw conclusion based on the data she

gathers, she should use inferential statistic method.

The type inferential statistic method that can be used is the confidence

interval.

From the data collected, the proportion of voters that view her favorably is

used at a given confidence level to determine the true proportion of voters

that view her favorably.

Therefore;

The statistical method that would be best for her to use is the

finding of the confidence interval which is a inferential statistic method.

Learn more here:

brainly.com/question/15873157

7 0

3 years ago

In a survey, 32% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result

adoni [48]

Answer:

$z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 proportion of pet owners contacted by telephone is NOT significantly lower than 0.32

Step-by-step explanation:

Data given and notation

n=210 represent the random sample taken

X=61 represent the number of pet owners contacted by telephone

$\hat p=\frac{61}{210}=0.29$ estimated proportion of pet owners contacted by telephone

$p_o=0.32$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportions is lower than 0.32.:

Null hypothesis: $p \geq 0.32$

Alternative hypothesis: $p < 0.32$

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932$

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

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

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 proportion of pet owners contacted by telephone is NOT significantly lower than 0.32

3 0

3 years ago

51/9 as a mixed number

Leona [35]

5 2/3 is 51/9 as a mixed nuber
51/9––>17/3=5 2/3

5 0

4 years ago

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