What is 0.02% of 40 million pounds?

Solve for X can you show me how to do this problem.

sergey [27]

Answer is 4.
Right angle= 90
50+ 10(4)= 90

7 0

3 years ago

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2. Which product is negative?

Paul [167]

Answer:

B) , C)

Step-by-step explanation:

Assuming that you are multiplying each of the numbers for each given answer.

Look at the amount of negative signs provided. Odd amounts of negative will always guarantee a product of a negative answer. Even amounts of negative will always guarantee a product of a positive answer.

In this case, C. -4 , -4 , -4 , 4 & B. -4 , -4, -4, 4 are your answers, for both only have 3 negative signs, which would make a negative answer.

~

8 0

3 years ago

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My number had a tens digit that is 8 more than the ones digit. Zero is not one of my digits

marin [14]

The tens digit is 8 more than the ones digit. There isn't any zeros.

The maximum digit is 9 because 10 would be 2 digits, 1 and 0.

So say the tens digits is 9.

9 _

The ones digit is 8 less so 9-8 = 1

9 1

There isn't any zeros which is correct.

The answer is 91.

4 0

3 years ago

What is the measure of YB?

pishuonlain [190]

The formula is
YC x YD = AY x YB

fill in what we know:

18 x 6 = 9 x YB

108 = 9 x YB

YB = 108 / 9

YB = 12

5 0

3 years ago

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

Let $\bar X$ = sample mean heights

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)

= 0.99492 or 99.5%

The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.

Therefore, 99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

3 0

3 years ago