How to solve <br> (2.31*10^-6)+(5.87*10^-4)

The height of males approximates a normal distribution. The average height of the players on an NBA team is 6.6 feet with a stan

lara [203]

D) 4.5 Standard Deviations Above the Mean

All you have to do is add .2 to 6.6 until you reach 7.5

6.8
7.0
7.2
7.4

Then you only need half of .2 to reach 7.5, so it's 4.5 standard deviations above the mean

6 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

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.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So $n = 6, s = \frac{106}{\sqrt{6}} = 43.27$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 523}{43.27}$

$Z = -2.61$

$Z = -2.61$ has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0

4 years ago

I dont know please help and show your work

Phoenix [80]

Area=6x10=60 feet as it is rectangle so all angles are 90° and oppossite sides are equal

4 0

3 years ago

HI can you help me with this question

dolphi86 [110]

Answer:

42 square centimeters

Step-by-step explanation:

The total of sides FA and AB will be half the perimeter of the rectangle, or 14 cm. Since AB is 9 cm, that means FA is 5 cm.

EF is the difference between EF and FA, so is 11 cm -5 cm = 6 cm.

FC is the same 9 cm length as AB, so we have enough information about the trapezoid to use the area formula:

A = (1/2)(b1 +b2)h

A = (1/2)(9 cm +5 cm)(6 cm) = 42 cm^2

The area of the trapezium is 42 cm^2.

6 0

4 years ago

5/8+3/4 / -2/3 - 5/6 =

m_a_m_a [10]

Answer:

-11/12

Step-by-step explanation:

Add 5/8+3/4 (6/8) which is 11/8 and then get -2/3 - 5/6 which simplifies to -3/2

Multiply -3/2 by 4 to get it to like terms

11/8 / -12/8

Multiply and then simplify.

If you need help, just comment!!

3 0

3 years ago

How to solve (2.31*10^-6)+(5.87*10^-4)

How to solve (2.3110^-6)+(5.8710^-4)