Given A(3, –7), under which reflection is A (3, 7)?

Mathematics

1 answer:

Paladinen [302]3 years ago

3 0

Answer:

reflection over the x-axis bc y changed

Step-by-step explanation:

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What is the axis of symmetry for the graph of y-4x=7-x^2

STALIN [3.7K]

It is symmetrical about the y axis

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4 years ago

Mrs. Khan is mailing 2 packages. The large package weighs 6 2 3 pounds. The small package weighs 3 1 2 pounds. Which statement B

vagabundo [1.1K]

Answer:

Mrs. Khan is mailing 2 packages. Subtract 623 – 312 best describe how to find out how much heavier the larger package is. Hence Option C is correct

Solution:

Given that

Mrs khan is mailing two packages.

Weight of larger package = 623 pounds

Weight of small package = 312 pounds

Now we need to find how much heavier larger package compare to smaller one is.

To get this if we remove weight of smaller package from larger package , we will get how much larger package is heavier.

That is Weight of larger package – Weight of smaller package = 623 pounds – 312 pounds

Hence option C) subtract 623 – 312 best describe how to find out how much heavier the larger package is.

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3 years ago

Read 2 more answers

A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of

liq [111]

Answer:

The probability of failure of both the bulbs is 0.4323.

Step-by-step explanation:

For an exponential distribution the distribution is given by

$f(x,\lambda )=\int_{0}^{x }\lambda e^{-\lambda x}dx$

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

$P(X)=\int_{0}^{x }\lambda _{X}e^{-\lambda _{X}x}dx$

Similarly for the bulb Y the distribution function is given by

$P(Y)=\int_{0}^{y }\lambda _{Y}e^{-\lambda _{Y}y}dy$

Thus the probability for both the bulbs to fail within 1500 hours is

$P(E)=\int_{0}^{1500}\int_{0}^{1500}\frac{1}{1400}e^{\frac{-x}{1400}}\cdot \frac{1}{1400}e^{\frac{-y}{1400}}dxdy\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}\int_{0}^{1500}e^{\frac{-x}{1400}}\cdot e^{\frac{-y}{1400}}dxdy)\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}e^{\frac{-x}{1400}}dx)\cdot (\int_{0}^{1500}e^{\frac{-y}{1400}}dy)\\\\P(E)=\frac{1}{1400^{2}}\times 920.473\times 920.473\\\\\therefore P(E)=0.4323$