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viktelen [127]
3 years ago
11

Please help me I am stuck on this question.

Mathematics
1 answer:
ladessa [460]3 years ago
3 0
I think it is 6 I hope you get it right good luck but I think that is the answer but it should be 6
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Plz help plzzzz plzzzzz
Licemer1 [7]
The answer is A. Even though it have a positive 3, the horizontal shift is opposite of what it is
4 0
3 years ago
Hannah was adding 5 plus -4. When explaining her work she said that the sum is 9 because you add the numbers and take the sign o
Scilla [17]

Answer:

The sum is 1

Step-by-step explanation:

When you add a negative number to a positive number, it is basically just subtracting the negative from the positive. To make it simple, 5 + -4 is just 5 - 4.

Hannah was wrong in saying you take the sign of the larger number because it would not matter if the first number was a million, the signs don't change based on the size of numbers.

6 0
4 years ago
If grandma lives in Palm Springs which is 100 miles away. How long does it take to get there if you drive a constant speed of 50
alexira [117]

Answer:

Step-by-step explanation:

i dont know unit rate 50 miles every hour

50/1

it take 2 hours to get to palmm springs

8 0
3 years ago
Is 99,782,829 divisible by 3?<br><br> HELP ME PLZ
Zinaida [17]

Answer:

Yes, answer is 33260943 .

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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}

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

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
3 years ago
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