Ask question

Ask question

All categories

2 years ago

9

Two competing gyms each offer childcare while parents work out. Gym A charges $9.00 per hour of childcare. Gym B charges $0.75 p

er 5 minutes of childcare. Which comparison of the childcare costs is accurate?

2 answers:

Solnce55 [7]2 years ago

6 0

They are the same amount. if you divide 60 by 5 it is 12 and when you multiply .75 by 12 it equals 9. so Both are $9.00 an hour.

andreev551 [17]2 years ago

6 0

Answer:

Gym B and Gym A charge the same hourly rate for the children.

Step-by-step explanation:

You might be interested in

The data represents the body mass index (BMI) values for 20 females. Construct a frequency distribution beginning with a lower

Papessa [141]

Answer:

Given:

Body mass index values:

17.7

29.4

19.2

27.5

33.5

25.6

22.1

44.9

26.5

18.3

22.4

32.4

24.9

28.6

37.7

26.1

21.8

21.2

30.7

21.4

Constructing a frequency distribution beginning with a lower class limit of 15.0 and use a class width of 6.0.

we have:

Body Mass Index____ Frequency

15.0 - 20.9__________3( values of 17.7, 18.3, & 19.2 are within this range)

21.0 to 26.9__________8 values are within this range)

27.0 - 32.9____________ 5 values

33.0 - 38.9____________ 2 values

39.0 - 44.9 _____________2 values

The frequency distribution is not a normal distribution. Here, although the frequencies start from the lowest, increases afterwards and then a decrease is recorded again, it is not normally distributed because it is not symmetric.

3 0

2 years ago

7. If W = 3Z, and Z is 30% of R, then W is what

dem82 [27]

Answer:

90%

Step-by-step explanation:

Let R = 100

Since, Z = 30% of R

Therefore, Z = 30% *100 = 30

Since, W = 3Z

Therefore, W = 3*30 = 90

Percentage of W of R

$= \frac{Value \: of \: W}{Value \: of \: R} \times 100\\\\=\frac{90}{100} \times 100\\\\Percentage \:of \:W \:of \:R =90\%$

7 0

2 years ago

Suppose X is the breaking strength (newtons) of a material, and X is normally distributed with

Veronika [31]

Answer:

a) 0.997 is the probability that the breaking strength is at least 772 newtons.

b) 0.974 is the probability that this material has a breaking strength of at least 772 but not more than 820

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 800 newtons

Standard Deviation, σ = 10 newtons

We are given that the distribution of breaking strength is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( breaking strength of at least 772 newtons)

$P(x \geq 772)$

$P( x \geq 772) = P( z \geq \displaystyle\frac{772 - 800}{10}) = P(z \geq -2.8)$

$= 1 - P(z$

Calculation the value from standard normal z table, we have,

$P(x \geq 772) = 1 - 0.003 = 0.997 = 99.7\%$

0.997 is the probability that the breaking strength is at least 772 newtons.

b) P( breaking strength of at least 772 but not more than 820)

$P(772 \leq x \leq 820) = P(\displaystyle\frac{772 - 800}{10} \leq z \leq \displaystyle\frac{820-800}{10}) = P(-2.8 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2.8)\\= 0.977 - 0.003 = 0.974 = 97.4\%$

$P(772 \leq x \leq 820) = 97.4\\%$

0.974 is the probability that this material has a breaking strength of at least 772 but not more than 820.

7 0

2 years ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

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

By the Central Limit Theorem

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

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

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

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c. The value exceeded by the mean time of the visitors with probability 0.01.

The 100 - 1 = 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

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

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

So 10.58 minutes.

6 0

2 years ago

Evelyn believes that if she flips a coin 480 times, it will land tails up exactly 240 times. What would you tell Evelyn about he

iragen [17]

Answer:

Step-by-step explanation:

I'd tell her that if (and only if) her prediction is correct, the probability of getting tails is 0.5. Actually, the probability of getting heads would be the same, 0.5.

In real life she would not necessarily get 240 tails out of 480 tosses, but the mean value of the number of tails would indeed be close to 0.5.

3 0

2 years ago