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Oduvanchick [21]
1 year ago
6

What is the answer to

Mathematics
2 answers:
marta [7]1 year ago
6 0

Answer:

224.50

<u>-225.00</u>

$0.50

- 50 cents

Yuki888 [10]1 year ago
4 0

Answer:_______________________________________________

-0.5

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A movie rental company offers a membership plan that costs $10/month and charges $2 to rent each movie. How many movies can a me
Brums [2.3K]

Answer:

See below

Step-by-step explanation:

linear equation: y=2x+10

x=cost per movie rented

10=$10 for the membership

24=2x+10    (24 being total you want to spend)

subtract 10 from each side

14=2x

divide by 2

7=x

He can rent no more than 7 movies if he want to spend $24 or less in the month

8 0
3 years ago
Please help ASAP I have no clue how to do this​
Salsk061 [2.6K]
It’s a triangle
s
r t

rs= 2x+10 and st= x-4

the sides are equal so equal them together then solve

2x+10=x-4
+ 4 +4
2x+14=x

you got x alone so x=2x+14

now you plug in 2x+14 in for the x in rs and st

rs= 2(2x+14)+10
4x+28+10
4x+38

st= (2x+14)-4
2x+14-4
2x-10
6 0
3 years ago
The Honda Accord was named the best midsized car for resale value for by the Kelley Blue Book (Kelley Blue Book website). The fi
Marianna [84]

Answer:

hello your question is incomplete below is the complete question

The Honda Accord was named the best midsized car for resale value for by the Kelley Blue Book (Kelley Blue Book website). The file AutoResale contains mileage, age, and selling price for a sample of 33 Honda Accords. Click on the datafile logo to reference the data. The estimated regression equation is Selling price = 20385.25049 - 0.03739 Mileage - 686.33668 age. Round your answers to the nearest dollar. a. Estimate the selling price of a four-year-old Honda Accord with mileage of 40,000 miles.

answer :  ≈ $16144

Step-by-step explanation:

<u>Determine the selling price of a four-year-old Honda Accord with mileage of </u><u>40,000</u><u> miles.</u>

Given data :

Age= 4 year, Mileage=40,000 miles

Hence the selling price

= 20385.25049 - 0.03739 Mileage - 686.33668 age

=  20385.25049 - 0.03739(40,000)  - 686.33668 ( 4 )

= $16144.32

 

4 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
4 years ago
I’m confused please help
mart [117]

Answer:

im guessing the answer is 43.7

Step-by-step explanation:

7 0
3 years ago
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