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Minchanka [31]
2 years ago
13

A candy package states that it now has 30% more candy. The original package had 5 ounces of candy. How much candy is in the new

package?​
Mathematics
1 answer:
Anastasy [175]2 years ago
7 0

Answer:

There are now 6.5 ounces in the new bag.

Step-by-step explanation:

To solve this question, take 30% of 5 and add it to the original amount of 5

30% = 0.30

5 * 0.30 = 1.5

5 + 1.5 = 6.5 ounces of candy.

Hope this helps!

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arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ 1^s^t statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

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

In this case we have:  \mu=5\ ft\sigma=0.4\ ft

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒2^n^d statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft  

Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒3^r^d statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒4^t^h statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

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