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Misha Larkins [42]
3 years ago
13

If a new phone screen is 19.68 cm2 larger than your current phone screen, with the dimensions 6.1 cm and 10.8 cm, then the new p

hone screen must have an area of ___, which is the product of 6.9 cm and ___. (A)6.9 cm,12.4 cm (B)65.88 cm2, 12.4 cm(C)85.56 cm2, 12.4 cm(D) 86 cm, 12 cm
Mathematics
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

(C)85.56 cm², 12.4 cm

Step-by-step explanation:

The Area of the screen of the new phone is :

19.68cm² + Area of the current phone screen size

The current phone screen size has dimensions: 6.1 cm and 10.8 cm,

Area of the current phone screen size = 6.1 cm × 10.8cm

Area of the current phone screen size = 65.88 cm²

Hence, The Area of the screen of the new phone is :

19.68cm² + Area of the current phone screen size

= 19.68cm² + 65.88cm²

= 85.56 cm²

The new phone screen must have an area of 85.56cm², which is the product of 6.9 cm and ___

The is calculated as:

85.56cm²/6.9cm

= 12.4cm

Therefore, the new phone screen must have an area of 85.56cm², which is the product of 6.9 cm and 12.4cm

Option C is correct

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There is a basketball with a diameter of 9 in. and a softball with a diameter of 4 in.. Find how many times greater the volume o
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The basketball has a volume of 113 inches .
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Simplify: (-1+4i)(-12-9i)
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-1 (-12 - 9i) + 4i (-12 - 9i)

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The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 ye
trasher [3.6K]

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 432, \sigma = 722, n = 40, s = \frac{722}{\sqrt{40}} = 114.16

What is the probability that the mean gain for the sample was between 250 and 500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.

So

X = 500

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

By the Central Limit Theorem

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

Z = \frac{500 - 432}{114.16}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257.

X = 250

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

By the Central Limit Theorem

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

Z = \frac{250 - 432}{114.16}

Z = -1.59

Z = -1.59 has a pvalue of 0.0559.

So there is a 0.7257 - 0.0559 = 0.6698 = 66.98% probability that the mean gain for the sample was between 250 and 500.

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Help me pls fast ASAP
tensa zangetsu [6.8K]

I think the answer is the last one.

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3 years ago
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