I have no clue on how to solve this

A total of 35 pints of blood were collected at a local blood drive how many quarts of blood were collected

Step2247 [10]

There was a total of 17.5 quarts of blood that was collected.

Hope this helps! :-)

7 0

3 years ago

Finding an angle measure

g100num [7]

Answer:

C. 50 degrees

Step-by-step explanation:

Angle EDG is 90 degrees

x = 10

5 multiplied by 10 is 50

4 multiplied by 10 is 40

40+50=90

3 0

3 years ago

Read 2 more answers

A box holds 42 pieces of red construction paper and 18 pieces of blue construction paper. A piece of paper is pulled out of the

dimulka [17.4K]

First problem :
probability to draw a red paper : 42/(42+18)=7/10

In 110 tries one can thus expect to draw 7/10*110=77 red papers

Second problem :
Since for 150 trials (which will give the least error) we have got 135 yellow balls, we can expect twice as much in 300=150*2 trials, which makes 270 yellow balls.

3 0

3 years ago

What is the explicit formula for the sequence? −1, 0, 1, 2, 3, . . .

rosijanka [135]

U r adding positive 1

8 0

4 years ago

Some research suggests that police officers are more likely to make an arrest in the presence of bystanders. If mentally disorde

zmey [24]

Answer:

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

$p_v =P(t_{(19)}>1.638)=0.0589$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=7.03$ represent the sample mean

$s=9.42$ represent the sample standard deviation

$n=20$ sample size

$\mu_o =3.58$ represent the value that we want to test

$\alpha=0.01$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 3.58 :

Null hypothesis: $\mu \leq 3.58$

Alternative hypothesis: $\mu > 3.58$

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".

Calculate the statistic

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

$t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

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

$p_v =P(t_{(19)}>1.638)=0.0589$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.

7 0

3 years ago