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3 years ago

The other two answer choices are 343 and 1,400! i’ve been stuck on these!!

Mathematics

1 answer:

kobusy [5.1K]3 years ago

3 0

Answer:

1,343

Step-by-step explanation:

divide 200 to 7 then multiply that to 12 and add 1000 and round

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X+k=w+v solve for x?

goldenfox [79]

Subtract k from both sides of problem

X+k-k = w+v-k

X= w+v-k

6 0

3 years ago

Read 2 more answers

Please solve and show the steps. 2/3z +3 =7

USPshnik [31]

Subtract 3 from both sides:

2/3z=4

divide by 2/3 on both sides (multiple 3/2 on both sides)

z= 6

4 0

3 years ago

Read 2 more answers

Help me please it would mean a lot

strojnjashka [21]

Answer:

2/6 = 1/3

Step-by-step explanation:

Make the denominators the same and minus them

7 0

3 years ago

Can someone help pleaseeeeeee

Anna007 [38]

Answer:

-5

Step-by-step explanation:

The parabolas equation is

(y - k)^2 = 4p(x - h)

Where h,k is the vertex

Substituting the vertex ad (2,-4)

(y - -4)^2 = 4p(x - 2)

(y +4)^2 = 4p(x - 2)

We need to find p from the other point they give us (-3,-3)

(-3 +4)^2 = 4p(-3 - 2)

1^2 = 4p (-5)

1 = -20p

Divide by -20

1/-20 = -20p/-20

-1/20 = p

Substituting back into the equation

(y +4)^2 = 4(-1/20)(x - 2)

Simplifying

(y +4)^2 = (-1/5)(x - 2)

FOILing

y^2 +8y +16 = -1/5x +2/5

Multiply by 5

5y^2 +40y +80 = -x +2

Subtract 2

5y^2 +40y +80-2 = -x +2-2

5y^2 +40y +78 = -x

Multiply by -1

-5y^2 -40y -78 = x

The coefficient of y^2 is -5

8 0

3 years ago

In one city, the probability that a person will pass his or her driving test on the first attempt is 0.68. Eleven people are sel

kolezko [41]

Answer:

0.0308 = 3.08% probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they pass the test on their first attempt, or they do not. Each person is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In one city, the probability that a person will pass his or her driving test on the first attempt is 0.68.

This means that $p = 0.68$

Eleven people are selected at random from among those taking their driving test for the first time.

This means that $n = 11$

What is the probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

This is

$P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{11,2}.(0.68)^{2}.(0.32)^{9} = 0.0009$

$P(X = 3) = C_{11,3}.(0.68)^{3}.(0.32)^{8} = 0.0057$

$P(X = 4) = C_{11,4}.(0.68)^{4}.(0.32)^{7} = 0.0242$

$P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0057 + 0.0242 = 0.0308$ .

0.0308 = 3.08% probability that among these 11 people, the number passing the test is between 2 and 4 inclusive

6 0

3 years ago