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2 years ago

Can you guys help me answer these!??

Mathematics

2 answers:

Margarita [4]2 years ago

8 0

3. 13

4. 66

5. 10

6. 16

alukav5142 [94]2 years ago

5 0

3. 16
4. 66
5. 20
6. 16

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Given f (x) = x +1 and g(x) = x², what is (gºf)(x)?

Gelneren [198K]

Answer:

Solution given:

f(x) = x +1 and g(x) = x²,

now

(gºf)(x) =g(fx)=g(x+1)=(x+1)²

B) (gºf)(x) = (x + 1)^2

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2 years ago

Find the zero of the function /(x) = 5x-3.

Bad White [126]

Answer:

3/5

Step-by-step explanation:

Set it equal to 0

5x-3=0

Then solve for x

5x=3

x=3/5

Check

5(3/5) - 3

3-3

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2 years ago

HELP PLEASE!!

Verizon [17]

did you ever get the answer?

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3 years ago

Read 2 more answers

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Alexxandr [17]

A number cube is a die (singular for dice). List all of the possible outcomes of a die (list numbers of 1 to 6), and next to those numbers, determine if they are even or odd.

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2 years ago

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

Note that: 1 is subtracted because it is a probability without replacement

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

2 years ago