What is the simplist form of 240/1000

Answers

Answer 1
Answer:

Divide by 100 to reduce it to 24/100

Now divide by 4 to get 6/25

The simplest form is 6/25


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Pablo folds a straw into a triangle with side lengths of 4x2 – 3 inches, 4x2 – 2 inches, and 4x2 – 1 inches. The perimeter of the triangle is 12x2 – 6 inches. If x = 1.5, what was the length of the straw before it was folded?___ inches

Answers

perimiter=legntho f straw=12x^2-6

if x=1.5 inches
12(1.5)^2-6=legnth
12(2.25)-6=
27-6=
21

legnth is 21 inches

Answer:

A

Step-by-step explanation:

Factor using ac-method and factor completely, 25x²-30x--40  (help)

Answers

25x^2-30x-40=\n 5(5x^2-6x-8)=\n 5(5x^2-10x+4x-8)=\n 5(5x(x-2)+4(x-2))=\n 5(5x+4)(x-2)
Answer :25x^2-30x-40 \n =5(5x^2-10x+4x-8) \n =5[5x(x-2)+4(x-2)] \n =5(x-2)(5x+4)

A cone has a base radius of 2 millimeters and a height of 24 millimeters.What is the volume of the cone?

A.
16π mm3

B.
384π mm3

C.
96π mm3 <-----

D.
32π mm3

Answers

The answer is D. 32? mm3. The volume of the cone is: V = 1/3 * π * r^2 * h, where r is the radius and h is the height. We know that: r = 2 mm and h = 24 mm. Therefore, the volume of the cone is: V = 1/3 * π * r^2 * h = 1/3 * π * 2^2 * 24 = 1/3 * π * 4 * 24 = 1/3 * π * 96 = 32 π mm^3.

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 standard deviation of 85. About how many scores greater than 600 would we expect to find?

Answers

\mathbb P(X>600)=\mathbb P\left((X-510)/(85)>(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*7500\approx10.86 scores above 600.

Final answer:

First, calculate the z-score for 600 using the provided mean and standard deviation. The result, approximately 1.059, indicates that it lies roughly one standard deviation above the mean. Approximately 15.87% of scores in a normal distribution are more than 1 standard deviation greater than the mean, so we would expect about 15.87% of 7500, or 1190 scores, to be greater than 600.

Explanation:

The question is asking about a phenomenon in statistics called the normal distribution, which is a type of continuous probability distribution often seen in natural parameters such as heights, weights, or scores, as in this case. To find how many scores are greater than 600, we first need to calculate the z-score, which measures how many standard deviations an element is from the mean.

The z-score formula is Z = (X - μ)/σ, where X is the score, μ is the mean, and σ is the standard deviation. In this case, X = 600, μ = 510, and σ = 85. So, Z = (600 - 510)/85 ≈ 1.059.

For a normal distribution, about 15.87% of the data falls more than 1 standard deviation greater than the mean. So, we would expect about 15.87% of 7500 scores to be greater than 600. That means, approximately 1190 scores would be greater than 600.

Learn more about Z-score and Normal Distribution here:

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I NEED HELP PLEASE ASAP!

Answers

Answer:

make a line with arrows at the end. On the line, make a hash mark and label it -13. on the hash mark, make a filled-in circle and color the number line to the left of the circle, including the arrow.

Step-by-step explanation:

This is how you show it. :)

A function is created to represent the balance on a credit card each month. What restrictions would be made to the range

Answers

Answer:

The range would include all real numbers.

Step-by-step explanation:

Consider the provided information.

Range of a function is the set of output values which a function can produce.

Let say we created a function to represent the balance on a credit card each month. Then the range will be the balance amount.

The balance amount can be a negative number as we are talking about credit card, also the balance can be zero or a positive number.

The credit card balance can be in decimals.

Thus, the balance can be any real number.

Hence, the range would include all real numbers.

It could include any numbers any real numbers.