Pls help! i’ll give brainly to the right answer

Alessandra's closet contains shelves that hold 12 short-sleeved shirts and 16 long-sleeved shirts. each shelf holds the same num

Arturiano [62]

12+16=28

28 divided by 7 = 4

<h2>the answer is B 4</h2>

6 0

3 years ago

PLEASE HELP I GIVE THANKS

xenn [34]

The answer is choice C
Plane D(choice C) because choice a and b are not lines(one letter cannot be a line and PBC and not connected). It's not d because e isn't a point

6 0

3 years ago

The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 ent

SOVA2 [1]

Answer:

$t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744$

$p_v =P(t_{(47)}>2.744)=0.0043$

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=2.1$ represent the mean

$s=1.01$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =1.7$ represent the value that we want to test

$\alpha=0.02$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 1.7, the system of hypothesis would be:

Null hypothesis: $\mu \leq 1.7$

Alternative hypothesis: $\mu > 1.7$

If we analyze the size for the sample is > 30 but 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".

Calculate the statistic

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

$t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744$

P-value

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

$df=n-1=48-1=47$

Since is a one side test the p value would be:

$p_v =P(t_{(47)}>2.744)=0.0043$

Conclusion

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.

4 0

3 years ago

Read 2 more answers

Help me Please Please Please Please Pleaseeeeeeee

Thepotemich [5.8K]

H = 151t - 16t²

The height of the ball when it return to the ground will be 0
0 = 151t - 16t²

The zero product property is that when two numbers are being multiplied and the product is 0, one of them must be equal to 0. Therefore, we can factorize this equation:
16t² - 151t = 0
t(16t - 151t) = 0
By the zero product property:
t = 0 or 16t - 151 = 0

So t = 0 or t = 9.44 seconds

The first solution is before he releases the ball and the second is when the ball comes back to the ground. Thus, the ball's air time is 9.44 seconds.

6 0

3 years ago

Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B

wariber [46]

Answer:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) is the probability that given that a student has a visa card, they also have a master card

p(A/B) is the probability that given a student has a master card, they also have a visa card

e. 0.35

f. 0.31

Step-by-step explanation:

a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)

=0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76

b. P(AnBnC')= P(AnB)-P(AnBnC)

=0.3-0.07= 0.23

c. P(B/A)= P(AnB)/P(A)

=0.3/O.6= 0.5

e. P((AnB)/C))= P((AnB)nC)/P(C)

=P(AnBnC)/P(C)

=0.07/0.2= 0.35

f. P((AUB)/C)= P((AUB)nC)/P(C)

=(P(AnC) U P(BnC))/P(C)

=(0.11+0.1)/0.2

=0.21/0.2 = 0.31

7 0

3 years ago