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Rudik [331]
3 years ago
12

Priya filled 5 jars, using a total of 7 ½ cups of strawberry jam. How many cups of jam are in each jar?

Mathematics
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

 1 1/2 cups of jam per jar

Step-by-step explanation:

Divide 7 1/2 by 5 to get 1 1/2 cups of jam per jar.

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The sum of a number and forty-six
ad-work [718]

\textsf{Hey there!}

\mathsf{\star \ The\ sum\ of\ a\ number\ \&\  forty-six}

\frak{Let's\ lable\ the\ keypoints\ so\ it\ can\ be\ easier\ to\ solve}

\bullet \ \textsf{The word \bf{\underline{SUM}}}\textsf{\ means\ add/addition}}

\bullet\textsf{ \underline{"A number"} is an unknown number so we can lable it as \bf{x}}

\bullet\textsf{ \underline{forty-six} is 46}

\textsf{The sum (+) does in the middle while \underline{x} \& \underline{46} will either be on left/right}

- \ \textsf{We can say that x is on your left}

-\  \textsf{  + can be in your middle}

- \ \textsf{\& lastly 46 can be on your right}

\boxed{\textsf{Thus, your answer SHOULD LOOK like: \boxed{\huge\text{\bf{x + 46}}}}}\checkmark

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4 0
3 years ago
What is 0.00078 in scientific notation?
frozen [14]
Answer: 7.8 x 10^-4
4 0
3 years ago
There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit
Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a \frac{3}{11} chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is \frac{8}{11}, and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is\frac{2}{11}, and the probability of picking the second red ball is \frac{1}{10} and the probability of picking the green ball is\frac{1}{9}. We want to multiply thm again, as per the multiplication rule like the last problem.

\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}

5 0
3 years ago
suppose you buy 158,000 home. you found out the bank offers 30 year loan at 6.5% APR. What will he your monthly payment
serg [7]

Answer: $998.67

Step-by-step explanation:

M = P[r(1+r)^n/((1+r)^n)-1)]

This does not include taxes, insurance, or PMI

8 0
3 years ago
If a ray bisects an angle, each newly formed angle measures 90 degrees.
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Answer:

A) Sometimes

A ) Sometimes is the correct answer because b) always and c) never is not possible

it cannot be always measured 90 degrees

3 0
3 years ago
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