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4 years ago

If y varies directly as x2 and y = 144 when x = 4, what is the value of y when x = -5?

1 answer:

muminat4 years ago

6 0

Y varies with x² times a constant. We will call the constant a
144=ax²
144 = a(4²)
144 = 16a
a = 9
Apply our equation y = 9x²
y = 9(-5)²
y = 9* 25
y=225
Answer is B

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Solve the following systems by substitution

Paladinen [302]

Answer:

The answer to your question is below

Step-by-step explanation:

Question 1

x = 5 Equation l

2x + y = 10 Equation ll

- Substitute Equation l in equation ll

2(5) + y = 10

y = 10 - 10

y = 0

- Solution (5, 0)

Question 2

x + 16y = 20 Equation l

x = 4y Equation ll

Substitute equation ll in equation l

4y + 16y = 20

20y = 20

y = 20/20

y = 1

-Find x

x = 4(1)

x = 4

-Solution

(4, 1)

Question 3

2x + 8y = 20 Equation l

x = 2 Equation ll

-Substitute equation ll in equation l

2(2) + 8y = 20

4 + 8y = 20

8y = 20 - 4

8y = 16

y = 16/8

y = 2

- Solution

(2, 2)

8 0

4 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

Point D is a fixed point on line PQ such that PD is =DQ, the cordinate of P and Q are (2,3) and (4,7)

Savatey [412]

Answer:

Step-by-step explanation:

hope this helps

5 0

3 years ago

1. 10(.15)=1.5

Fed [463]

Answer:

6.14125(0.15) = 0.9211875 (below 1)

6.14125 - 0.92118 = 5.22007

Step-by-step explanation:

Given data

1. 10(.15)=1.5

2. 10-1.5=8.5

3. 8.5 (.15)=1.275

4. 8.5-1.275=7.225

continuation the sequence

5) 7.225 (0.15) = 1.08375

6) 7.225 - 1.08375 = 6.14125

7) 6.14125(0.15) = 0.9211875 (below -one)

8 ) 6.14125 - 0.9211875 = 5.2200625 (get number 5)

9) 5.2200625(0.15) = 0.783009

10) 5.2200625 - 0.783009 = 4.4370532

3 0

3 years ago

Use the function below to find F(1). F(t) = 2•1/2^3t

stiv31 [10]

Answer:

D) 1/4

Step-by-step explanation:

To solve this, first lets figure out what this variable is. We know we must multiply by t. Well, looking at it, we know that t is one. So lets plug that in.

2*1/2^3(1)

Now we need to figure out what the exponent is that 2 will be raised to. To find that, find out what 3 times one is.

2*1/2^3

Now we need to solve the exponent.

2*1/8

Remember, 2 cubed is eight.

Now we need to find out what 2*1/8 is. To solve this more easily, I am going to turn two into a fraction. This will give us 2/1*1/8. Now we cross multiply, multiplying the numerators by the numerators, and the denominators by the denominators. Remember, the numerator is the top number, and the denominator is the bottom number.

If we do that, we get:

2/8

If we simplify it, we get 1/4. So the asnwer is D) 1/4

6 0

3 years ago