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4 years ago

Talia uses 10 strawberries and 4 oranges to make 8 servings of fruit salad. How many strawberries for 12 servings?

1 answer:

8 0

For 8 servings, Talia uses strawberries = 10
For 1, it is = 10/8

Now, for 12 servings, it would be: 10/8 * 12 = 10/2 * 3 = 5 * 3 = 15

In short, Your Answer would be 15 strawberries

Hope this helps!

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5 bricks weigh 26 1/4 how much do 1 brick weigh

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Answer:

5 1/4

Step-by-step explanation:

5 1/4 + 5 1/4 + 5 1/4 + 5 1/4 + 5 1/4 = 6 1/4

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3 years ago

Emmalissa goes to the movies with 4 friends. They will each pay an equal share for the movie tickets. Ozzie volunteered to put i

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3 years ago

A jar contains 0.65 liter of lime juice and 0.40 liter of orange juice. Meg poured 0.35 liter of cranberry juice into the jar. S

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Part B: 1.3L left and basic maths

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3 years ago

Read 2 more answers

The results of Accounting Principals’ latest Workonomix survey indicate the average American worker spends $1092 on coffee annua

Nataly_w [17]

Answer:

Group 18-34 years old

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

Group 35-44 years old

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

Group 45 and older

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

According to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average, 47.74% probability that a person between 35 and 44 consume more than the average and 50% probability that a person older than 45 consume more than the average.

Step-by-step explanation:

The mean for each sample is

$\bar x=\frac{\sum_{k=1}^{10}x_k}{10}$

where the $x_k$ are the data corresponding to each group

The variance is

$s^2=\frac{\sum_{k=1}^{10}(\bar x-x_k)^2}{9}$

and the standard deviation is s, the square root of the variance.

Group 18-34 years old

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

Group 35-44 years old

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

Group 45 and older

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

Let's compare these averages against the general media established of $1,092 by using the corresponding z-scores

$z=\frac{\bar x-\mu}{s/\sqrt{n}}$

where

$\bar x$ = mean of the sample

$\mu$ = established average

s = standard deviation of the sample

n = size of the sample

z-score of Group 18-34 years old

$z=\frac{1041.625-1092}{696.635/\sqrt{10}}=-0.2286$

The area under the normal curve N(0;1) between -0.2286 and 0 is 0.0904. So according to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average.

z-score of Group 35-44 years old

$z=\frac{1359-1092}{422.5491/\sqrt{10}}=2.0019$

The area under the normal curve N(0;1) between 0 and 2.0019 is 0.4774. So according to the sample there is 47.74% probability that a person between 35 and 44 consume more than the average.

z-score of Group 45 and older

$z=\frac{1414.375-1092}{135.2489/\sqrt{10}}=7.5375$

The area under the normal curve N(0;1) between 0 and 7.5375 is 0.5. So according to the sample there is 50% probability that a person older than 45 consume more than the average.

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4 years ago

Each day, mr. ling has a boy and a girl summarize the assignment for the day. there are 8 boys and 9 girls in his group. in how

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8x9=72

In 72 ways. Hope this helps.

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3 years ago