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3 years ago

Write an equation in point-slope form

Mathematics

2 answers:

Slav-nsk [51]3 years ago

5 0

Answer:

Step-by-step explanation:

stepladder [879]3 years ago

4 0

I believe is C because the formula is

Y = mx + b

Since your m value is -6/5 and your x value is -3, and y value is -7

Then your answer would be

Y - 7 = -6/5 (x - 3)

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Round to the given place: 256,035 (thousands)

mash [69]

So,

When rounding to the nearest thousand, look at the number in the hundreds place. If it is greater than or equal to 5, round up. If not, round down.

256,035

The hundreds place is less than 5, so we round down.

256,035 --> 256,000

5 0

3 years ago

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35% of the birds in a sanctuary are hummingbirds. If there are 420 birds in the sanctuary

Arlecino [84]

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3 years ago

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A= 3x3x3x3x3x5x7x7x7

olga nikolaevna [1]

Answer:

LCM(57624, 416745) = 23,337,720 or LCM = 23337720

Step-by-step explanation:

List all prime factors for each number.

Prime Factorization of 57624 is:

2 x 2 x 2 x 3 x 7 x 7 x 7 x 7 => 23 x 31 x 74

Prime Factorization of 416745 is:

3 x 3 x 3 x 3 x 3 x 5 x 7 x 7 x 7 => 35 x 51 x 73

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

The new superset list is

2, 2, 2, 3, 3, 3, 3, 3, 5, 7, 7, 7, 7

Multiply these factors together to find the LCM.

LCM = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 7 x 7 x 7 x 7 = 23337720

In exponential form:

LCM = 23 x 35 x 51 x 74 = 23337720

LCM = 23337720

Therefore,

LCM(57624, 416745) = 23,337,720

3 0

2 years ago

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali

Softa [21]

Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

$W =$ Wednesday

$T =$ Thursday

$F =$ Friday

$S =$ Saturday

Their corresponding probabilities are

$P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1$

Since $Y$ = number of days beyond Wednesday that it takes for both magazines to arrive(so possible $Y$ values are 0, 1, 2 or 3)

The possible number of outcomes are therefore $4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)$

The values associated for each of the outcomes are as follows:

$Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3$

The probability mass function of $Y$ is,

$P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19$