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3 years ago

Write an expression for the following. "The sum of a number and 12 times 4

Mathematics

1 answer:

andriy [413]3 years ago

7 0

Answer:

4(x + 12)

Step-by-step explanation:

The sum of a number and 12 indicates addition. Then, it says that this sum will be multiplied by 4. So, we can use x to represent this random number.

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PLEASE HELP WILL MARK BRAINLIEST<br> -SEE ATTACHMENT-

yuradex [85]

ANSWER

$(x - 5) ^{5} = {x}^{5} - 25 {x}^{4} + 250 {x}^{3} - 1250 {x}^{2} + 3125x- 3125$

EXPLANATION

From the Pascal's triangle,

The coefficients are :

1, 5, 10,10, 5,1

Because of the negative sign, the expansion will alternate sign.

$(x - y) ^{5} = {x}^{5} - 5 {x}^{4} y + 10 {x}^{3} {y}^{2} - 10 {x}^{2} {y}^{3} + 5x {y}^{4} - {y}^{5}$

We now put y=5 into the expansion to get,

$(x - 5) ^{5} = {x}^{5} - 5 {x}^{4} (5) + 10 {x}^{3} ( {5}^{2}) - 10 {x}^{2} ({5})^{3} + 5x( {5})^{4} - {5}^{5}$

$(x - 5) ^{5} = {x}^{5} - 25 {x}^{4} + 250 {x}^{3} - 1250 {x}^{2} + 3125x- 3125$

3 0

3 years ago

Read 2 more answers

A software developer wants to know how many new computer games people buy each year. Assume a previous study found the standard

Marrrta [24]

Answer:

The minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 95% level of confidence is n=567.

Step-by-step explanation:

We have to calculate the minimum sample size n needed to have a margin of error below 0.14.

The critical value of z for a 95% confidence interval is z=1.96.

To do that, we use the margin of error formula in function of n:

$MOE=\dfrac{z\cdot \sigma}{\sqrt{n}}\\\\\\n=\left(\dfrac{z\cdot \sigma}{MOE}\right)^2=\left(\dfrac{1.96\cdot 1.7}{0.14}\right)^2=(23.8)^2=566.42\approx 567$

The minimum sample size to have this margin of error is n = 567.

3 0

2 years ago

Jayda is watching a movie that lasts 1 hour and 37 minutes. She has watched 52 minutes so far. How many minutes are left in the

Svetlanka [38]

Answer:

45 minutes

Step-by-step explanation:

97-52 = 45

5 0

3 years ago

Which set of ordered pairs gives the polygon shown?

makvit [3.9K]

The correct answer is B.

Explanation
Since each interval mark is 1/4 of a unit, we will write this as 0.25. For the first point, 4 interval marks to the left of the y-axis makes it a negative number; 4(0.25) = 1; this makes the x-coordinate of this point -1. 2 interval marks above the x-axis makes it positive; 2(0.25) = 0.5; this makes the y-coordinate 0.5. This makes the first ordered pair (-1, 0.5).

The second point is on the y-axis. This makes the x-coordinate 0. It is 5 intervals above the x-axis; this makes it positive. 5(0.25) = 1.25 will be the y-coordinate, making the point (0, 1.25).

The third point is 3 intervals to the right of the y-axis; this makes it positive, and 3(0.25)=0.75 for the x-coordinate. It is 3 intervals below the x-axis; this makes it negative, and 3(0.25) = 0.75, making the y-coordinate -0.75. This puts the third point at (0.75, -0.75).

7 0

3 years ago

In the university library elevator there is a sign indicating a 16-person limit as well as a weight limit of 2750 pounds. Suppos

Alona [7]

Answer:

0.039 = 3.9% probability that the random sample of 16 people in the elevator will exceed the weight limit

Step-by-step explanation:

To solve this question, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .