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Serggg [28]
1 year ago
15

The function f(x) = –(x – 20)(x – 100) represents a company’s monthly profit as a function of x, the number of purchase orders r

eceived. Which number of purchase orders will generate the greatest profit?
Mathematics
2 answers:
ad-work [718]1 year ago
8 0
The correct answer is choice A. 20
puteri [66]1 year ago
5 0

Answer:

Correct answer is 60

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A set of data has a mean of 56.1. The data follows a normal distribution curve and has a standard deviation of 8.2. Find the pro
Whitepunk [10]
Given that mean=56.1 and standard deviation=8.2, P(x>67.5) will be found as follows:
The z-score is given by:
z=(x-μ)/σ
thus the z-score will be given by:
z=(67.5-56.1)/8.2
z=11.4/8.2
z=1.39
thus
P(z=1.39)=0.9177
thus:
P(x>67.5)=1-P(z>0.9177)
=1-0.9177
=0.0823
Answer: A. 0.0823
7 0
2 years ago
Read 2 more answers
The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If
frutty [35]
For this case we have the following equation:
 w = 0.9* \sqrt{E}
 Where,
 w: The weight of a spring in pounds
 E: the energy stored by the spring in joules.
 Substituting values we have:
 w = 0.9* \sqrt{12}
 Making the corresponding calculation:
 w=3.12
 Answer:
 
the approximate weight of the spring in pounds is:
 
w=3.12
4 0
2 years ago
Read 2 more answers
A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average
larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: \mu \leq  94

Alternate Hypothesis: \mu > 94

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Using the values, we get:

t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

\alpha =0.05

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0
2 years ago
The assistant manager of a surf shop estimates that 65% of customers will make a purchase. Part A: How many customers should a s
Ivan

Answer:

A) 2

B) 0.239

Step-by-step explanation:

Part A) Expected value of a binomial distribution is the number of trials times the probability of success.

X = np

Given X = 1 and p = 0.65:

1 = 0.65n

n = 1.54

Rounding up, a salesperson should expect 2 customers until he finds a customer that makes a purchase.

Part B) Use binomial probability.

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

Given n = 3, r = 1, p = 0.65, q = 0.35.

P = ₃C₁ 0.65¹ 0.35³⁻¹

P ≈ 0.239

Or, using binompdf function in a calculator:

binompdf(n, p, r)

= binompdf(3, 0.65, 1)

≈ 0.239

3 0
2 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. The box plot compares the monthly average
vovangra [49]

Hello,

The median of the temperatures at Springwood is 86. (median is at the middle).

the median of the temperatures at Meadows is 73. (median is at the middle).

The (IQR) interquartile range of the temperatures at Meadows is 12. (You have to subtract both of the IQR left and right 80 - 68 = 12).

The (IQR) interquartile range of the temperatures at Springwood is 14. (You have to subtract both of the IQR left and right 91 - 77 = 14).

the difference of the medians as a multiple of their average interquartile range is 79.5 is average median and 13 average IQR.

5 0
1 year ago
Read 2 more answers
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