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Mila [183]
3 years ago
7

Henry had 2.5 lb of apples for a fruit salad. The recipe says that 5 lb of apples are needed for the salad. Henry went to the st

ore and bought an additional 1.5 lb of apples. Did Henry get a reasonable amount of apples from the store? A. Yes, Henry got a reasonable amount of apples because he only needed 1.5 lb of apples (2.5 – 1 = 1.5) B. No, Henry did not get a reasonable amount of apples because he needed a total of 7.5 pounds of apples (2.5 + 5 = 7.5). C. No, Henry did not get a reasonable amount of apples because he needed 2.5 more pounds of apples (2.5 + 2.5 = 5).
Mathematics
2 answers:
svetoff [14.1K]3 years ago
8 0
<span> C. No, Henry did not get a reasonable amount of apples because he needed 2.5 more pounds of apples (2.5 + 2.5 = 5).

</span>
NeTakaya3 years ago
4 0
The answer is C, no he did not get a reasonable amount of apples because he needed a total of 5 lbs, while he only had 4 lbs. 
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Step-by-step explanation:

6x - 3y

=> x = 5, y = -1 Substitute in the above equation,

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=> 6(5) + 3

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Please help me with number ten and twelve
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Answer: 93,\ 34^{\circ}

Step-by-step explanation:

Given

For figure (i) both triangles are congruent

\therefore 8x-27=4x+33\\\Rightarrow 8x-4x=33+27\\\Rightarrow 4x=60\\\Rightarrow x=15

\Rightarrow PQ=8x-27\\\Rightarrow PQ=93

For figure (ii)

Two triangles WXZ and WZY are congruent

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Guys, please help me with this question.
Elodia [21]

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2

Step-by-step explanation:

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= (x² - 5x²) + (xy + xy) + (- 3y² - y² )

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6 0
3 years ago
A recipe calls for 3 cups of sugar and 9 cups of water. How many cups of water should be used with 12 cups of sugar?
Yuki888 [10]

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According to the National Vital Statistics, full-term babies' birth weights are Normally distributed with a mean of 7.5 pounds a
Sav [38]

Answer:

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 7.5, \sigma = 1.1

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So

X = 8.6

Z = \frac{X - \mu}{\sigma}

Z = \frac{8.6 - 7.5}{1.1}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 6.4

Z = \frac{X - \mu}{\sigma}

Z = \frac{6.4 - 7.5}{1.1}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

6 0
3 years ago
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