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3 years ago

8

Maddie has 4 cups of maple syrup. She uses 1/3 of the syrup to make maple candies. Then, she uses 2/3 of a cup of syrup to make

a maple glaze for a ham. How much maple syrup does Maddie have left?

2 answers:

Reptile [31]3 years ago

8 0

Answer:

3 cups

Step-by-step explanation:

1/3 + 2/3 = 1 cup so you would subract 1 from 4 which is 3

hope this helps:)

AnnyKZ [126]3 years ago

8 0

Answer: 2 cups of syrup left

======================================================

Explanation:

She uses 1/3 of the syrup, leaving 2/3 of the syrup left after making the maple candies.

2/3 of 4 = (2/3)*4 = 8/3 cups are left over after making the candies

-----------

Then we subtract off 2/3 since it says "she uses 2/3 of a cup...for a ham"

Ultimately we have 8/3-2/3 = 6/3 = 2 cups left over

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ACCORDING TO QUESTION:-
Cucumbers = 6 for $2
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For 9 cucumbers = as 6 cucumbers is for $2 then 9 will be of $3........1️⃣

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Express 5/7 as a decimal. (Round to the nearest thousandths place)

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Answer:

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Step-by-step explanation:

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3 years ago

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34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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2 years ago

Mary had a 3-inch piece of ribbon. She cut it in half. How long were they new pieces

allochka39001 [22]

Answer:

1 and a half inches long

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What is the probability that from a normal 52 card deck, you randomly draw a 5, and then without replacement, you select a Queen

Mamont248 [21]

Answer:

1 / 663

Step-by-step explanation:

First, let's find the total number of ways you can pick 2 cards from the deck. This is 52 * 51 = 2652 because there are 52 cards available for your first pick, and after you pick one, you'll have 51 left for your second pick.

There are 4 5's (one for every suite) and only 1 Queen of Hearts in a deck of cards. Therefore, the total number of successful outcomes will be 4 * 1 = 4.

The probability of picking a 5 and then a Queen of Hearts is 4 / 2652 =

1 / 663. Hope this helps!

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