Timothy explains to his brother that the model y=4mt has a constant of variation of 4 and is an example of _____ variation.

Mathematics

1 answer:

Lunna [17]2 years ago

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Profit:p=r-c;solve for c

eduard

If you would like to solve p = r - c for c, you can do this using the following steps:

p = r - c /+c
p + c = r - c + c
p + c = r /-p
p + c - p = r - p
c = r - p

The correct result would be c = r - p.

5 0

2 years ago

Read 2 more answers

Twenty-five juniors from a University volunteer to allow their Math GRE test scores to be used in a study. These 25 juniors had

fgiga [73]

Let's start first by writing down the given:
σ = 100
sample mean = 450
sample size = 25

These information, plus the fact that we know that the population is approximately normally distributed, would tell us that we can use the normal distribution curve in analyzing the problem.

A confidence interval of the mean is just a range statistically estimated to contain the population mean. For a 90% confidence interval, we would look at the Z-table and see where 90% of the data falls. We'll notice that it will fall within 1.645 standard deviations of the mean.

Next, we look for the standard error of the mean. This will have a formula
$error= \frac{standard deviation of population}{ \sqrt{N} }$

The standard error would just therefore be equal to
$\frac{100}{ \sqrt{25} }= \frac{100}{5} =20$

Lastly, we just get the product of the standard error and 1.645 and add it to 450 for the maximum value and subtract it to 450 for the minimum value.
$450-20(1.645)=417.1$
$450+20(1.645)=482.9$

ANSWER: 417.1<μ<482.9

8 0

2 years ago

What is the probability that a point chosen at random in the triangle is also in the blue square? A blue square inside of an uns

Dmitrij [34]

Answer:

One-third

Step-by-step explanation:

we know that

The probability that a point chosen at random in the triangle is also in the blue square is equal to divide the area of the square by the area of triangle

step 1

Determine the area of triangle

$A=\frac{1}{2}(9)(6)=27\ in^2$