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3 years ago

5

If m 1 + m 5 = 100 find the measure of 2

1 answer:

Mariana [72]3 years ago

6 0

If the question is to answer m2, than m2≈33.34 roughly

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In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 point

aivan3 [116]

Step-by-step explanation:

Below is an attachment containing the solution.

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3 years ago

A rate is a ratio? ( sometimes, always or never true?)

Brums [2.3K]

Answer:

The answer is never true

Step-by-step explanation:

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2 years ago

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How to do this? pls don't ignore

oksano4ka [1.4K]

So basically you just need to do the Pythagorean theorem to find the sides of the shaded rectangle, with c being the length of the side.

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2 years ago

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Simplify.Your answer should contain only positive exponents.

zhannawk [14.2K]

Step-by-step explanation:

1) $r^3r^3$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

$=r^{3+3}$

$=r^6$

2) $3k^3k^3$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

$=3k^{3+3}$

$=3k^6$

3) $\left(3b^2\right)^3$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n$

$=3^3\left(b^2\right)^3$

$=27\left(b^2\right)^3$

$=27b^6$

4) $\left(2m^3\right)^3$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n$

$=2^3\left(m^3\right)^3$

$=8\left(m^3\right)^3$

$=8m^{3\cdot \:3}$

$=8m^9$

5) $\frac{3r^3}{r^2}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}$

$=3r^{3-2}$

$=3r^1$

$=3r$

3 0

2 years ago

What is the probability that the number of “HEADs” on these four coins is equal to 3? (4 points)

slavikrds [6]

Answer:

a. $\frac{1}{4}$

Step-by-step explanation:

We are asked to find the probability of getting 3 heads on 4 flips.

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Since we know that flipping a fair coin has 2 equally likely possible outcomes, so flipping four coins will have $2*2*2*2=16$ possible outcomes.

Sample space of possible outcomes.

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,

THHH, THHT, THTH,THTT, TTHH, TTHT, TTTH, TTTT.

We can see that there are 4 favorable outcomes of getting heads.

$\text{Probability of getting 3 heads}=\frac{4}{16}$

$\text{Probability of getting 3 heads}=\frac{1}{4}$

Therefore, the probability of getting 3 heads on 4 coins will be $\frac{1}{4}$ and option a is the correct choice.

5 0

3 years ago