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hey! please help with this!!! Write all the 12 answer of measurements. If you can’t do it and you think it’s a lot you don’t hav

jarptica [38.1K]

Answer:

1) 1 and 4

2) 1 and 2

3) 1 and 5

4) 2 and 3

5) 130

6) 90

7) 25

8) 115

9) 70

10) 115

11) 40

12) 105

Step-by-step explanation:

Hope this isn't spam! Lol

4 0

3 years ago

Put these in order from greatest to least.

nikklg [1K]

86%, 3/4, 0.682, 0.67, 5/8, 0.41

^ I converted the above to decimals first and then arranged them that way.

1/20 = 0.05 x 100 = 5%

7 0

3 years ago

Read 2 more answers

What is the determinant of the coefficient matrix of the system?<br> –158<br> –117<br> 65<br> 117

tatuchka [14]

Answer:

C

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A random sample of 180 microbiology students were asked how many science classes he or she was enrolled in August 1990. The resu

frutty [35]

Answer:

$z=\frac{1.83-1.94}{\sqrt{\frac{1.48^2}{180}+\frac{1.62^2}{180}}}}=-0.673$

$p_v =2*P(z$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the two means

Step-by-step explanation:

Data given and notation

$\bar X_{1}=1.83$ represent the mean in 1990

$\bar X_{2}=1.94$ represent the mean for 2005

$s_{1}=1.48$ represent the sample deviation for 1990

$s_{2}=1.62$ represent the sample standard deviation for 2005

$n_{1}=180$ sample size for 1990

$n_{2}=180$ sample size for 2005

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis: $\mu_{1}=\mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

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

$z=\frac{1.83-1.94}{\sqrt{\frac{1.48^2}{180}+\frac{1.62^2}{180}}}}=-0.673$

What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

$p_v =2*P(z$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the two means

8 0

4 years ago

Add. −5/12+(−3/4) Enter your answer in the box as a fraction or mixed number in simplest form.

brilliants [131]

The answer is 1/2 cuz of simplyfing

5 0

3 years ago