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miv72 [106K]
3 years ago
9

Jonah makes a snack mix to take with him on vacation. His favorite snack mix is 5 parts raisins and 4 parts peanuts. He wants to

take 27 cups of snack mix. How many cups of raisins and peanuts does he need
Mathematics
1 answer:
Arisa [49]3 years ago
3 0
(3) 5 parts raisins
(3) 4 parts peanuts
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saw5 [17]

Answer:

the answer should be 152.6 ! just moving the decimal over to the left!!

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For every blue strip you have, you need two yellow strips. You have 12 blue strips. How many yellow strips do you need ?
Katen [24]

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24 yellow strips

Step-by-step explanation:

since you need two yellow for every blue, you need to double the total of blue to get the number of yellow that are needed.

7 0
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Solve.
mina [271]
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4y\ \textgreater \ -8-16 \\
4y\ \textgreater \ -24 \\
y\ \textgreater \ \frac{-24}{4} \\
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Solve for the missing variable. Round to the nearest hundredth if necessary.
vagabundo [1.1K]

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3 0
3 years ago
birth weights in norway are normally distributed with a mean of 3570 g and a standard deviation of 500g. if the hospital officia
Svetradugi [14.3K]

Answer:

2630 g

Step-by-step explanation:

From the given information:

Given that:

mean (μ) = 3750 g

Standard deviation (σ) = 500

Suppose the hospital officials demand special treatment with a percentage of  lightest 3% (0.03) for newborn babies;

Then, the weight of birth that differentiates the babies that needed special treatment from those that do not can be computed as follows;

P(Z < z₁) = 0.03

Using the Excel Formula; =NORMSINV(0.03) = -1.88

z₁ = - 1.88

Using the test statistics z₁ formula:

z_1 = \dfrac{X-\mu}{\sigma}

-1.88 = \dfrac{X-3570}{500}

By cross multiply, we have:

-1.88 × 500 = X - 3570

-940 = X - 3570

-X = -3570 + 940

-X = -2630

X = 2630 g

Hence, 2630 g is the required weight of birth   that differentiates the babies that needed special treatment from those that do not

6 0
2 years ago
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