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3 years ago

The product of a constant factor 15 and a 2-term factor x+4

1 answer:

6 0

The product of 15 and $x+4$ is $15(x+4) = 15x+60$ . This is a linear binomial.

$4x^3-2x-9-(2x^3)$ is a cubic trinomial. In your original question, I'm not sure what the $5x+3$ means.

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Help me with this pls

vladimir2022 [97]

90 counterclockwise hope I’m right

8 0

2 years ago

Read 2 more answers

A football field measures 120 yards long and 54 yards wide. We know that the team can fit 26 players in 5 yards by 8 yards box.

wariber [46]

Answer:

4,212 people

Step-by-step explanation:

We are given;

Dimensions of the football field are 120 yards by 54 yards
26 players can fit in a box measuring 5 yards by 8 yards

We are required to determine the number of people that can fit in the football field.

We are going to get the area of the field first;

Area = length × Width

Area of the field = 120 yards × 54 yards

= 6480 square yards

Then we get the area of the box

Area of the box = 5 yards × 8 yards

= 40 square yards

We then determine the number of boxes that can fit in the football field

Number of boxes = Area of the foot ball field ÷ Area of one box

= 6480 sq yards ÷ 40 sq yards

= 162 boxes

But, 26 people can fit in one box

Therefore,

Total number of people that can fit in the field is given by multiplying the number of boxes that can fit in the field by the number of people in a box

That is, Number of people = 26 players × 162 boxes

= 4,212 people

Therefore, the number of people that can fit in the football field is 4,212 people.

6 0

3 years ago

Evaluate C=2 pi r for r=8 <br> A. C=10 pi <br> B. C=16 pi<br> C. C=28 pi<br> D. C=16

Brrunno [24]

Answer:

If `r` and `R` and the respective radii of the smaller and the bigger semi-circles then the area of the shaded portion in the given figure is: (FIGURE) `pir^2\ s qdotu n i t s` (b) `piR^2-pir^2\ s qdotu n i t s` (c) `piR^2+pir^2\ s qdotu n i t s` (d) `piR^2\ s qdotu n i t s`

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

How to solve this problem I don’t get it at all

Zarrin [17]

Use a calculator to help u and u will get your answer

8 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 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 $\frac{\sigma}{\sqrt{n}}$ .

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.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

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

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

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

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

3 years ago