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3 years ago

Use elimination to solve this system of equations. X-4y=-11 and 5x-7y=-10

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Hopefully this helps you start to understand the steps to take in elimination

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If the Price Elasticity of Demand for a product is -.8, and the product currently sells for $100 per unit, how much will demand

egoroff_w [7]

Answer: Percentage change in quantity would decrease by 12.5%.

Step-by-step explanation:

Since we have given that

Price elasticity of demand = -0.8

Price of each unit = $100

After change, price of each unit becomes = $110

% change in quantity would be

$\dfrac{New-old}{old}\times 100=\dfrac{110-100}{100}\times 100=10\%$

As we know the relation between price, quantity and elasticity.

$e_D=\dfrac{\text{\% change in quantity}}{\text{\%change in price}}\\\\-0.8=\dfrac{10\%}{x}\\\\x=\dfrac{10}{-0.8}\\\\x=-12.5\%$

Hence, Percentage change in quantity would decrease by 12.5%.

8 0

3 years ago

CD is the perpendicular bisector AB. G is the midpoint of AB. points E and F lie on CD. which pair of line segments must be cong

Ugo [173]

AG and GB must be congruent and CG and GD must be congruent as well

6 0

3 years ago

Read 2 more answers

Newton's first law of motion applies only to objects that are already moving. O True O False

larisa [96]

Answer:

false

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

6m + 2 - 4m = help with this please

devlian [24]

The answer is: 2(m+1)

First combine like terms.
we do 6m-4m which is 2m
2m+2=?

Second we factor. We notice that both terms have 2.
2(m+1)

by distributive law, the 2 is distributed to m and 1.

8 0

2 years ago

A researcher determines that students are active about 60 + 12 (M + SD) minutes per day. Assuming these data are normally distri

rjkz [21]

Answer:

The correct option is (b).

Step-by-step explanation:

If X $\sim$ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z $\sim$ N (0, 1).

The distribution of these z-variate is known as the standard normal distribution.

The mean and standard deviation of the active minutes of students is:

μ = 60 minutes

σ = 12 minutes

Compute the z-score for the student being active 48 minutes as follows:

$Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0$

Thus, the z-score for the student being active 48 minutes is -1.0.

The correct option is (b).

4 0

3 years ago