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KIM [24]
4 years ago
12

Misha and Dontavious were asked to solve the following problem using linear programming. A local school district can operate a l

arge bus for $300 per day and a small bus for $200 per day. The district needs to arrange transportation for at least 360 students for a field trip and has enough chaperones to have one chaperone per bus for up to 7 buses. Each large bus can hold 60 students and each small bus can hold 45 students. What is the minimum cost for the transportation for the field trip?
Mathematics
2 answers:
bearhunter [10]4 years ago
4 0

Answer:

did you get the answer?

Step-by-step explanation:

AURORKA [14]4 years ago
4 0

Answer:

1700 for the bus

Step-by-step explanation:

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BRAINLISET BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST
Vinvika [58]

Answer:

x = -15/2

Step-by-step explanation:

For this problem, we will simply use equation properties to solve for x.

2x - 5 = -20

2x - 5 + 5 = -20 + 5

2x = -15  ( Add positive 5 to both sides )

2x * (1/2) = -15 * (1/2)

x = -15/2  ( Multiply both sides by 1/2)

Hence, the solution to x is -15 / 2.

Cheers.

4 0
3 years ago
Read 2 more answers
(x ÷ 8) + 23 – 4 = 36
kifflom [539]

Answer:

(x ÷ 8) + 23 – 4 = 36

x = 136

Step-by-step explanation:

3 0
2 years ago
Mrs. Adams deposited $12,000 into an account that earns an annual simple interest rate of 3.25%. She makes no other deposits or
lisabon 2012 [21]

Answer:

<em>Mrs. Adams will earn $3,120 of interest at the end of year 8.</em>

Step-by-step explanation:

<u>Simple Interest</u>

In simple interest, the money earns interest at a fixed rate, assuming no new money is coming in or out of the account.

We can calculate the interests earned by an investment of value A in a period of time t, at an interest rate r with the formula:

I=A.r.t

Mrs. Adams deposited an amount of A=$12,000 into an account that earns an annual simple interest rate of r=3.25%. We must find the interest earned in t=8 years. The interest rate is converted to decimal as:

r=3.25/100=0.0325

The interest is then calculated:

I=12,000\cdot 0.0325\cdot 8=3,120

Mrs. Adams will earn $3,120 of interest at the end of year 8.

7 0
3 years ago
A group of tenth grade students responded to a survey that was asked which Math course they were currently enrolled in. The surv
ale4655 [162]

Answer:

C.

Step-by-step explanation:

The bottom-right most cell tells us that the total number of students that responded to the survey is 310 students.

To find the answer, we can go through each choice.

A. Females taking Geometry

Row 1 Column 2 tells us that 53 females are taking geometry. 53/310 is about 17%.

B. Females taking Algebra II.

Row 1 Colume 3 tells us that 62 females are taking Algebra II. 62/310 is 20%.

C. Males taking Geometry.

Row 2 Colume 2 tells us that 59 males are taking Geometry. 59/310 is about 19%. Choice C is correct.

D. Males taking Algebra I.

44 out of the total 310 respondents are male and is taking Algebra I. 44/310 is about 14%.

4 0
3 years ago
The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0
2 years ago
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