How do I write 770,070 in expanded form

The information below describes a data plan for a storage device.

Vesna [10]

Answer

A) C=5+2d

Step-by-step-explanation

3 0

3 years ago

Find the slope of the line

Mashcka [7]

3/2 it’s three up and two to the right

6 0

3 years ago

Simplify the expression. - | -13 | =

a_sh-v [17]

Since the -13 has the absolute value signs next to it, the 13 becomes a positive, but there is a negative outside the absolute value, and the negative sign and the absolute value sign cancelled previously, so the expressions is now
-13
The absolute value signs cancel out a negative only inside the signs, not outside, so the negative sign in the outside keeps.

5 0

4 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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 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.

The mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

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

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

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

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0

3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

oksano4ka [1.4K]

Answer:

a.) dx3x² + 2

Use the properties of integrals

That's

integral 3x² + integral 2

= 3x^2+1/3 + 2x + c

= 3x³/3 + 2x + c

= x³ + 2x + C

where C is the constant of integration

b.) x³ + 2x

Use the properties of integrals

That's

integral x³ + integral 2x

= x^3+1/4 + 2x^1+1/2

= x⁴/4 + 2x²/2 + c

= x⁴/4 + x² + C

c.) dx6x 5 + 5

Use the properties of integrals

That's

integral 6x^5 + integral 5

= 6x^5+1/6 + 5x

= 6x^6/6 + 5x

= x^6 + 5x + C

d.) x^6 + 5x

integral x^6 + integral 5x

= x^6+1/7 + 5x^1+1/2

= x^7/7 + 5/2x² + C

Hope this helps

8 0

3 years ago