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3 years ago

The histogram represents a data distribution with uniform class widths 1.

Mathematics

2 answers:

kondor19780726 [428]3 years ago

8 0

Answer:

Step-by-step explanation:

aliina [53]3 years ago

4 0

I think it’s b... I don’t but i tried if it is wrong I’m sorry if it was right your welcome

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Complete the solution of the equation. Find the

forsale [732]

Answer:

y = -2

Step-by-step explanation:

Put x as -19 and solve for y.

2(-19) - 5y = -28

Multiply.

-38 - 5y = -28

Add 38 on both sides of the equation.

-5y = -28 + 38

-5y = 10

Divide both sides by -5.

y = 10/-5

y = -2

3 0

4 years ago

Read 2 more answers

Caroline has 10 marbles. One half of them are blue. How many marbles are blue?

Viktor [21]

Answer: 5

One half of 10 is 10/2 or 1/2(10)
10/2 = 5
1/2(10) = 5

6 0

3 years ago

Read 2 more answers

You roll a six-sided number cube (die). What is the BEST answer for the probability that the number rolled is between 1 and 6, i

posledela

Answer: It is A certain.

Step-by-step explanation:

Because all the numbers on a six-sided cube is between 1 and 6 so it is certain or 100/100 that the number will land on a number between 1 and 6.

4 0

3 years ago

Suppose you have a bag of red, blue, and yellow marbles. If the probability of picking a red marble is start fraction 1 over 6 e

Alona [7]

Answer: B. one-third

Step-by-step explanation:

Probability of picking a red marble, P(R) = 1/6

Probability of picking a yellow marble, P(Y) = 1/2

Therefore, the probability of picking a blue marble will be:

= 1 - (P(R) + P(Y))

= 1 - (1/6 + 1/2)

= 1 - (4/6).

= 1 - 2/3

= 1/3

The probability of picking a blue marble will be one third

8 0

3 years ago

I need to know how to do the answer and why that’s the anwer

Pavel [41]

Ok so the answer is
$4 {x}^{2}$

so we multiply numbers that have the same variable

$2 {x}^{3} \: \: \times {2x}^{ - 1} \\ \\$
multiply 2 x 2

$2 \times 2 = 4$
then multiply the exponents

$x ^{3} \times x ^{ - 1}$
multipling like this will look like addition.
${x}^{3} \times {x}^{ - 1} = {x}^{2}$
so in other words you could do it like this too
${x}^{3} + {x}^{ - 1} = {x}^{2}$
now add the exponent with 4
$4 + {x}^{2} = 4x ^{2}$

now for the last part
${y}^{3} + {y}^{ - 3} = 0$
because -3 and 3 added together make 0
$0 + 4 {x}^{2} = 4 {x}^{2}$
I hope this help(it looks complicated but it's very easy and I hope I helped you out)

6 0

3 years ago