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3 years ago

Find the value of y and the measure of each angle labeled in the figure. (2y – 15) (Y+ 3)

Mathematics

1 answer:

vredina [299]3 years ago

5 0

Answer:

$\frac{443 + 53 \\ 1 =khskk \frac{ = = = 5 + + | = = = {222}^{2} | }{?} 5x}{?}$

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A bag contains 5 blue marbles, 2 green marbles, and 3 yellow marbles. What is the probability of choosing one green marble and t

Vesna [10]

Answer:

Step-by-step explanation:

Total number of marbles = 5 + 2 + 3 = 10

Probability of choosing 1 green marble = 2/10

Probability of choosing 1 yellow marble = 3/10

Notice (and this is important) that the denominator didn't change. Why?

Because you replaced the first marble into the bag. That word replacement is critical in a problem of this nature. There is the term non replacement which means that the second draw would have a denominator of 9.

So what is the probability of P(green then yellow)?

P(green then yellow) = 3/10 * 2/10 = 6/100

Answer: P(green then yellow) = 3/50 because 6 / 100 reduces.

4 0

3 years ago

I don’t know the answer pls help I’ll give you brainliest!! 19 points.

gladu [14]

Answer:

it is A i think i am a nub tho

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

$4,700 is invested in an account earning 6.2% interest (APR), compounded

pickupchik [31]

Answer:

(5 x (-20)) * (100 x (0.7))

4 0

2 years ago

Find the upper limit for the zeros of the function P(x)=4x^4+8x^3-7x^2-21x-9

pishuonlain [190]

Answer:

OPTION B - 2

Step-by-step explanation:

What is the upper limit for the zeros of the function P(x) = 4x^4 + 8x^3 - 7x^2 - 21x - 9. Ans: 2 is an upper limit. Use synthetic division. and the remainder are all positive, 2 is an upper limit.

3 0

3 years ago

A bag contains 6 green counters, 4 blue counters and 2 red counters. Two counters are drawn from the bag at random without repla

Svetach [21]

Answer:

24.24%

Step-by-step explanation:

In other words we need to find the probability of getting one blue counter and another non-blue counter in the two picks. Based on the stats provided, there are a total of 12 counters (6 + 4 + 2), out of which only 4 are blue. This means that the probability for the first counter chosen being blue is 4/12

Since we do not replace the counter, we now have a total of 11 counters. Since the second counter cannot be blue, then we have 8 possible choices. This means that the probability of the second counter not being blue is 8/11. Now we need to multiply these two probabilities together to calculate the probability of choosing only one blue counter and one non-blue counter in two picks.

$\frac{4}{12} * \frac{8}{11} = \frac{32}{132}$ or 0.2424 or 24.24%

8 0

3 years ago