Estimate 396*37 then find the product

Vaselesa [24]

I'm thinking you would have to add the probabilities that you choose a boy with you choose someone in the science club minus a boy in the science club. You can't count these people twice. I'm getting 59% (11+10-2)/32

7 0

3 years ago

Jonas needs at least $27 to buy some school supplies. He already has $21. Which graph best represents all the possible dollar am

Ahat [919]

If Jonas has $21, and he needs $27, he is going to have to save $6 or more
Therefore, your answer is 1.

Hint: If it is Greater Than or Less Than, the circle is closed. If it has Greater Than or Equal To or Less Than or Equal To, the Circle is closed.

Hope this Helps!

6 0

4 years ago

Read 2 more answers

Find the proceeds for the simple discount note.

lutik1710 [3]

Given:
Face Value $15,000
Discount rate 6.5%
Term 70 days

Discount for a simple discount note:
Discount = Maturity Value * Discount Rate * Term of Note
Discount = 15,000 * 0.065 * 70/360
Discount = 189.58

Proceeds from Simple Discount note:
Proceeds = Maturity Value - Discount
Proceeds = 15,000 - 189.58
Proceeds = 14,810.42

The proceeds for the simple discount note is $14,810.42. Choice C.

5 0

3 years ago

Read 2 more answers

Hey guys I need help with this

bulgar [2K]

Answer:

use the reverse Pythagorean therm BC= 4 DE= 4.8

Step-by-step explanation:

The square root of 5(squared) - 3(squared) = 4

the square root of 8(squared) - 6.4(squared) = 4.8

7 0

3 years ago

For the following discrete random variable X with probability distribution:

Black_prince [1.1K]

Answer:

(a) The probability distribution is shown in the attachment.

(b) The value of E (Y) is 7.85.

(c) The value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d) The value of P (Y ≤ 2) is 0.60.

(e) Verified that the value of E (Y) is 7.85.

Step-by-step explanation:

(a)

The random variable Y is defined as: $Y=3X^{2}-2X+1$

For X = {0, 1, 2, 3} the value of Y are:

$X=0;\ Y=3\times(0)^{2}-2\times(0)+1 =1$

$X=1;\ Y=3\times(1)^{2}-2\times(1)+1 =2$

$X=2;\ Y=3\times(2)^{2}-2\times(2)+1 =9$

$X=3;\ Y=3\times(3)^{2}-2\times(3)+1 =22$

The probability of Y for different values are as follows:

P (Y = 1) = P (X = 0) = 0.20

P (Y = 2) = P (X = 1) = 0.40

P (Y = 9) = P (X = 2) = 0.15

P (Y = 22) = P (X = 3) = 0.25

The probability distribution of Y is shown below.

(b)

The expected value of a random variable using the probability distribution table is:

$E(U)=\sum[u\times P(U=u)]$

Compute the expected value of Y as follows:

$E(Y)=\sum [y\times P(Y=y)]\\=(1\times0.20)+(2\times0.40)+(9\times0.15)+(22\times0.25)\\=7.85$

Thus, the value of E (Y) is 7.85.

(c)

Compute the expected value of X as follows:

$E(X)=\sum [x\times P(X=x)]\\=(0\times0.20)+(1\times0.40)+(2\times0.15)+(3\times0.25)\\=1.45$

Compute the expected value of X² as follows:

$E(X^{2})=\sum [x^{2}\times P(X=x)]\\=(0^{2}\times0.20)+(1^{2}\times0.40)+(2^{2}\times0.15)+(3^{2}\times0.25)\\=3.25$

Thus, the value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d)

Compute the value of P (Y ≤ 2) as follows:

$P (Y\leq 2)=P(Y=1)+P(Y=2)=0.20+0.40=0.60$

Thus, the value of P (Y ≤ 2) is 0.60.

(e)

The value of E (Y) is 7.85.

$E(Y)=E(3X^{2}-2X+1)=3E(X^{2})-2E(X)+1$

Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).

$E(Y)=3E(X^{2})-2E(X)+1\\=(3\times 3.25)-(2\times1.45)+1\\=7.85$

Hence verified.

3 0

3 years ago