I need some help with this

Please helpppp

Annette [7]

Answer:

1. 41

2. 7.857 or 8

3. 21

4.

3 0

3 years ago

Read 2 more answers

What substance makes up 75 to 90% of every cell in the human body?

tensa zangetsu [6.8K]

Your answer is D, water.

8 0

2 years ago

a) The mean age of graduate students at a University is at most 31 y ears with a standard deviation of two years. A random sampl

kari74 [83]

Answer:

A.)

H0: μ ≤ 31

H1: μ > 31

B.)

H0: μ ≥ 16

H1: μ < 16

C.)

Right tailed test

D.)

If Pvalue is less than or equal to α ; we reject the Null

Step-by-step explanation:

The significance level , α = 0.01

The Pvalue = 0.0264

The decision region :

Reject the null if :

Pvalue < α

0.0264 > 0.01

Since Pvalue is greater than α ; then, we fail to reject the Null ;

Then there is no significant evidence that the mean graduate age is more Than 31.

B.)

H0: μ ≥ 16

H1: μ < 16

Null Fluid contains 16

Alternative hypothesis, fluid contains less than 16

One sample t - test

C.)

Null hypothesis :

H0 : μ ≤ 12

. The direction of the sign in the alternative hypothesis signifies the type of test or tht opposite direction of the sign in the null hypothesis.

Hence, this is a right tailed test ; Alternative hypothesis, H1 : μ > 12

d.)

If Pvalue is less than or equal to α ; we reject the Null.

3 0

2 years ago

Match the missing term with the perfect square trinomial 9X^2+?+25<br> 49<br> 6x<br> 36<br> 30x

Igoryamba

9x^2+t+25=(ax+b)^2

9x^2+t+25=a^2*x^2+2abx+b^2 so

a^2=9 so a=3

b^2=25 so b=5 then

t=2ab using a and b from above

t=2*3*5=30

So the missing term is 30x

4 0

3 years ago

The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o

timurjin [86]

Answer:

a) Null hypothesis: $\mu = 90000$

Alternative hypothesis: $\mu \neq 90000$

b) $p_v =2*P(t_{(24)}$

c) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation 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 values obtained are:

$\bar X=85272$ represent the mean annual salary for the sample

$s=11039.23$ represent the sample standard deviation for the sample

$n=25$ sample size

$\mu_o =90000$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:

Null hypothesis: $\mu = 90000$

Alternative hypothesis: $\mu \neq 90000$

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".

Part b: Calculate the statistic

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

$t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(24)}$

Part c: Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

6 0

3 years ago