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3 years ago

I need not just the answer but the work written out

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

6 0

Answer:

A(3) = -6 + (3 - 1) (5)

-6 + (2)(5)

-6 + (10)

A(4) = -6 + (4 - 1) (5)

-6 + (3)(5)

-6 + 15

A(10) = -6 + (10 - 1) (5)

-6 + (9)(5)

-6 + 45

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, 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 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:

$\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33$

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

$2.33 = \frac{X - 117}{4.33}$

$X - 117 = 2.33*4.33$

$X = 127.1$

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0

4 years ago

Building a is 160160 feet shorter than building

Serjik [45]

I assume the heights are 160 ft and 1480 ft.

The two heights are unknown, so we will use variable h to help solve the problem.
The shorter building, building A, has height h.
Since building A is shorter by 160 ft, then building B is taller by 160 ft, so the height of building B is h + 160.

Now we add our two heights to find the total height.

h + h + 160 is the total height.
We can write it as 2h + 160

We are told the total height is 1480 ft, so we let 2h + 160 equal 1480, and we have an equation.

2h + 160 = 1480

Subtract 160 from both sides

2h = 1320

Divide both sides by 2

h = 660

h + 160 = 820

Building A measures 660 ft.
building B measures 820 ft.

7 0

3 years ago

Use the list method to write "the perfect square integers between 1 and 80 inclusive".

Len [333]

<span> all of our square numbers from 1 to 80 are 1, 4, 9, 16, 25, 36, 49, 64. </span>

5 0

3 years ago

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A rectangular garden is 45 feet wide and 70 feet long. On a blueprint, the width is 9 inches. Identify the length on the bluepri

sweet [91]

A rectangular garden is 45 ft wide and 70 ft long. On a blueprint, the width is 9 in. Identify the length on the blueprint.

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