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4 years ago

Is 2 DVDs / $36.75 equal to $18.375?

Mathematics

2 answers:

Colt1911 [192]4 years ago

8 0

2/36.75 = 1/18.375

yes

otez555 [7]4 years ago

3 0

Just subtract those two numbers because then you will know if it is =2 that

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HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP!

11Alexandr11 [23.1K]

Answer:

Hey!

Your answer should be

Step-by-step explanation:

To find the BASE: 8 x 4 = 32cm^2

To find the angled TOP FACE: 10 x 4 = 40 cm^2

To find the TRIANGLES: (6x8)/2 x 2 = 48 cm^2

To find the REAR FACE: 6 x 4 = 24 cm^2

TOTAL: 32 + 40 + 48 + 24

SUFACE AREA: 144cm^2

Hope this helps!

7 0

3 years ago

Francisco purchased a 2010 model sedan for $24,999. The dealership offered him a $199/month

beks73 [17]

900 because look 51 * 71 = 9187 so then just after simplify

8 0

3 years ago

Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about 6.9. Use the criterion tha

zavuch27 [327]

Answer:

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$\mu = 80, \sigma = 6.9$

Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean

This means that z-scores higher than 2.5 or lower than -2.5 are considered unusual.

Would it be unusual to have more than 120 successes out of 200 trials

We have to find the Z-score of X = 120.

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

$Z = \frac{120 - 80}{6.9}$