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Nitella [24]
3 years ago
12

The scatter plot below shows the relationship between the outside temperature and the number of cups of hot chocolate sold at an

event. Which statement describes the data?
A. There is no association between the outside temperature, in degrees Fahrenheit, and the number of cups of hot chocolate sold.
B. There is a nonlinear association between the outside temperature, in degrees Fahrenheit, and the number of cups of hot chocolate sold
C. There is a positive linear association between the outside temperature, in degrees Fahrenheit, and the number of cups of hot chocolate sold.
D. There is a negative linear association between the outside temperature, in degrees Fahrenheit, and the number of cups of hot chocolate sold.
Mathematics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

B

Step-by-step explanation:

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According to creditcard , the mean outstanding credit card debt of college undergraduate was $3173 in 2010. A researcher believe
Mamont248 [21]

Answer:

a. The null and alternative hypothesis can be written as:

H_0: \mu=3173\\\\H_a:\mu< 3173

b. A Type I error is made when a true null hypothesis is rejected. In this case, it would happen if it is concluded that the actual mean outstanding credit card debt of college undergraduate is significantly less than $3173, when in fact it does not.

A Type II error is made when a false null hypothesis is failed to be rejected. In this case, the actual mean outstanding credit card debt of college undergraduate is in fact less than $3173, but the test concludes there is no enough evidence to claim that.

Step-by-step explanation:

We have a prior study of the mean outstanding credit card debt of college undergraduate that states that it was $3173 in 2010.

A researcher believes that this amount has decreased since then.

Then, he has to perform a hypothesis test where the null hypothesis states that the mean is still $3173 and an alternative hypothesis that states that the actual credit card debt is significantly smaller than $3173.

The null and alternative hypothesis can be written as:

H_0: \mu=3173\\\\H_a:\mu< 3173

3 0
3 years ago
4 1/5 tons = how many ponds
elena55 [62]

Answer:

Step-by-step explanation:

21/5*2000=  21*400= 8,400 pounds

5 0
2 years ago
Read 2 more answers
I need help on question 2 i need the simplified answer and the restrictions
Anastaziya [24]

Answer:

x cannot be 5 or -3.

\frac{x+5}{x+3}

Step-by-step explanation:

The restrictions for a fraction is that the bottom cannot be 0.

So if we find when the bottom is 0 we have found the values that x cannot be.

Let's solve x^2-2x-15=0.

Since the coefficient of x^2 is 1 all we have to do is find two numbers whose product is -15 and whose sum is -2.

Those numbers are -5 and 3 since (-5)(3)=-15 and (-5)+(3)=-2.

So the factored form of the equation is:

(x-5)(x+3)=0

This means either x-5=0 or x+3=0.

We do have to solve both.

x-5=0 can be solved by adding 5 on both sides.

x-5+5=0+5

x+0=0+5

x=5

x+3=0 can be solved by subtracting 3 on both sides.

x+3-3=0-3

x+0=0-3

x=-3

So x can be any number except x=5 or x=-3.

We already factored the bottom as (x-5)(x+3).

The top is a difference of squares, x^2-a^2,

which can be factored as (x-a)(x+a).

So the top factors as (x-5)(x+5).

The fraction can then be written as:

\frac{x^2-25}{x^2-2x-15}=\frac{(x-5)(x+5)}{(x-5)(x+3)}

This can further simplified assuming x is not 5 we can write it as \frac{x+5}{x+3}.  I canceled the common factor of (x-5).

5 0
3 years ago
PLEASE HELP ME WITH BOTH OF THESE PROBLEMS THIS IS DUE SOON PLEASE HELP ME
iogann1982 [59]

Answer:

Step-by-step explanation:

4.

a)

P = 2width + 2length

8x = 2x + 2 length; 8x-2x = 2 length; 6x=2 length

length = 3x

b)

sides= 1/4 of 3x = 3x/4

perimeter of square = 4* side = 4*3x/4 = 3x

c)

x= 4

P rectangle =8x= 8*4= 32

P square = 3x = 3*4 = 12

32-12 = 20 cm greater is the P rectangle than P of square

d)

A rectangle = 2x*x = 2*4*4 = 32

A square =(3x/4)² = (3*4/4)² = 9

32-9 = 23 cm² greater is the A rectangle than A of square

5.

4Comic+1 Nonfiction

4comic = 8y so 1 comic = 2y

1 nonfiction = (3+7y)+ 2y

a)

1 nonfiction = 7y+2y+3 = 9y+3

b)

total= 8y+9y+3 = 17y+3

if y=4

total = 17*4 +3  = $71

7 0
3 years ago
Read 2 more answers
The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical
Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{36 - 38.4}{0.8538}

Z = -2.81

Z = -2.81 has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

Z = \frac{X - \mu}{s}

Z = \frac{41 - 38.4}{0.8538}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08

So

a)

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

b)

Z = \frac{X - \mu}{s}

Z = \frac{36 - 38.4}{1.08}

Z = -2.22

Z = -2.22 has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

Z = \frac{X - \mu}{s}

Z = \frac{41 - 38.4}{1.08}

Z = 2.41

Z = 2.41 has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

4 0
3 years ago
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