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1 year ago

713.49 written in a standard form

Mathematics

1 answer:

Vikki [24]1 year ago

7 0

Answer: one hundred seven thirteen and forty-nine hundreds

Step-by-step explanation: Standard form is a way of writing down very large or very small numbers easily. 103 = 1000, so 4 × 103 = 4000 . So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form.

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Which expression is equivalent to 12r + 8p –34r – 2p?

ELEN [110]

Answer:

It is C -14r+6p

Step-by-step explanation:

If you want the explanation just tell me

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2 years ago

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The width of a rectangle is the length minus 2 units. The area of the rectangle is 35 units. What is the width, in units, of the

Andrews [41]

Answer: width = 5 units

Step-by-step explanation:

Let L represent the length of the rectangle.

The width of a rectangle is the length minus 2 units. It means that the width of the rectangle is (L - 2) units.

The formula for determining the area of a rectangle is expressed as

Area = length × width

The area of the rectangle is 35 units. It means that

L(L - 2) = 35

L² - 2L = 35

L² - 2L - 35 = 0

L² + 5L - 7L - 35 = 0

L(L + 5) - 7(L + 5) = 0

L - 7 = 0 or L + 5 = 0

L = 7 or L = - 5

Since the length cannot be negative, then

L = 7 units

Width = 7 - 2 = 5 units

6 0

2 years ago

Consider a set of 7500 scores on a national test whose score is known to be distributed normally with a mean of 510 and a standa

german

$\mathbb P(X>600)=\mathbb P\left(\dfrac{X-510}{85}>\dfrac{600-510}{85}\right)=\mathbb P(Z>1.059)\approx0.145$

So approximately 14.5% of the scores are higher than 600. This means in a sample of 7500, one could expect to see $0.145\times7500\approx10.86$ scores above 600.

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1 year ago

A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s

amid [387]

Answer:

The correct conclusion is:

"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."

Step-by-step explanation:

A doctor is measuring the average height of male students at a large college.

The doctor measures the heights, in inches, of a sample of 40 male students from the baseball team.

Using this data, the doctor calculates the 95% confidence interval (63.5, 74.4).

The following conclusions is valid:

"The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches."

Since we know that the confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.

For the given case, the confidence level is 95% and the corresponding confidence interval is (63.5, 74.4) which represents the true mean of heights for male students at the college where the doctor measured heights.

Therefore, it is valid to conclude that the doctor is 95% confident that the mean height of male students at the college is within the interval of (63.5, 74.4).

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2 years ago

5(n+2)-8=2n what s the solution for x??

PIT_PIT [208]

5n+10-8=2n
5n+2=2n
-3n=2
n=-2/3

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2 years ago

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