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3 years ago

5. Jamal Willis deposits $475 in a savings account. The account pays an annual interest of 5

Mathematics

1 answer:

Makovka662 [10]3 years ago

8 0

Answer emily dont need a answer she doing thuff

Step-by-step explanation:

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Ethan buys a bag of cookies that contains 6 chocolate chip cookies, 5 peanut butter cookies, 7 sugar cookies and 7 oatmeal cooki

Inessa05 [86]

Answer:7/25 × 5/24

Step-by-step explanation:

7 oatmeal 25 all together

Then 5 peanut butter 24 left since he ate one

7 0

3 years ago

Explain how to convert numbers from standard form to scientific.

sweet [91]

Answer:

Start by moving the decimal place in the number until you have a coefficient between 1 and 10. The number of places to the left that you had to move the decimal point is the exponent. so for example, 0.000345 would be 3.45(10)^4

Step-by-step explanation:

Hope that helps!

8 0

3 years ago

A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading

valkas [14]

Answer:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

$p_v =2*P(t_{11}$

Step-by-step explanation:

1) Data given and notation

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

The results obtained are:

$\bar X=98.375$ represent the sample mean

$s=0.6.109$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =100$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

4) P-value

First we need to find the degrees of freedom for the statistic given by:

$df=n-1=12-1=11$

Since is a two sided test the p value would given by:

$p_v =2*P(t_{11}$

5) Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.