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2 years ago

EEEEEEOOOOOOOOOOGGGGGG plz answer my eog test

Mathematics

2 answers:

iVinArrow [24]2 years ago

6 0

Answer:

1.A 2.B 3.C 4.b 5.A 6. D

Step-by-step explanation:

Hope this helps

andrey2020 [161]2 years ago

3 0

Answer:

1.B, 2.D, 3.D, 4.C, 5.B, 6.D

Step-by-step explanation:

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. Is (3, 6) a solution to this system of equations? 2x + 7y = 48 5x + 3y = 51 .

pychu [463]

No (3,6) is the solution for the first equation only

6 0

2 years ago

PLEASE HELP 50 point will give brainlyest

Nutka1998 [239]

When I did it I got 140296.11541307

5 0

2 years ago

Read 2 more answers

10% of the downtown residents of a town in the Plains live in mobile homes. Based on data from past tornadoes, if a tornado goes

tia_tia [17]

Answer:

The probability a person will not live in a mobile home if they do not die as a result of the tornado is 0.9265.

Step-by-step explanation:

Denote the events as follows:

X = a person lives in a mobile home.

Y = a person dies as a result of tornado.

Given:

$P(X)=0.10\\P(Y|X) = 0.30\\P(Y|X^{c})=0.02\\$

Compute the value of $P(Y^{c}|X^{c})$ as follows:

$P(Y^{c}|X^{c})=1-P(Y|X^{c})\\=1-0.02\\=0.98$

Compute the probability that a person dies as a result of tornado as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})\\=(0.30\times0.10)+(0.02\times(1-0.10))\\=0.03+0.018\\=0.048$

Then the probability that a person does not dies as a result of tornado is:

$P(Y^{c})=1-P(Y)\\=1-0.048\\=0.952$

Compute the value of $P(X^{c}|Y^{c})$ as follows:

$P(X^{c}|Y^{c})=\frac{P(Y^{c}|X^{c})P(X^{c})}{P(Y^{c})}=\frac{0.98\times(1-0.10)}{0.952}=0.9265$

Thus, the probability a person will not live in a mobile home if they do not die as a result of the tornado is 0.9265.

5 0

3 years ago

I need to know how to get the answer

alukav5142 [94]

Step 1: Add 9 to both sides
Step 2: Combine like terms (notice the -9 and 9 cancel out)
Step 3: Flip the inequality

-16 > x - 9
-16 + 9 > x - 9 + 9 ... apply step 1
-7 > x ... apply step 2
x < -7 ... apply step 3

The final answer is x < -7
Any value smaller than -7 will work as a solution

8 0

3 years ago

Suppose that a candy company makes a candy bar whose weight is supposed to be 50 grams, but in fact, the weight varies from bar

siniylev [52]

Answer:

Step-by-step explanation:

Given that the weight of a candy bar X is N(mean = 50 gm, 2 = sigma)

Packs of 4 bars

n = sample size = 4

Std error of sample = sigma/sqrt n = 1

P(Mean diff > 4 gms) = P(Z>4/Se ) = P(Z>4)

=0.0000

Hence it is an impossible event

Option A

(A) It is extremely unlikely for this to occur; the probability is very close to 0.

7 0

3 years ago