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Solnce55 [7]
2 years ago
11

Using the pie chart below, if 300 students were polled on their favorite ice cream flavor, how many more students preferred stra

wberry ice cream than vanilla ice cream?

Mathematics
1 answer:
Paladinen [302]2 years ago
3 0
Since there are 300 students you have to use proportions to solve the percents. First for strawberry you need to set up a proportion such as x/300=30/100 to show you are trying to find 30%of 300. Then you must cross multiply so 300 times 30 and 100 times x. Then you have 9,000 and 100, so you divide 9,000 by 100 and then you get 90 which is the portion of kids who liked strawberry flavored ice cream. You should do the exact same thing for vanilla ice cream like this: x/300=29/100 but since you’re finding 29% you need to put a 29 over 100. Then you cross multiply and do 300 times 29 and do x times 100. Which makes you get 8,700 and 100, and then you divide 8,700 by 100 and get 87 for vanilla ice cream. Lastly you need to subtract 87 from 90 since it asked how many more students liked strawberry than vanilla: 90-87=

Can you please mark branliest? Thank you!
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SCORPION-xisa [38]

Think of the 13-ft length of the ladder as the hypotenuse of a right triangle. Represent the horiz. distance from foot of ladder to base of tree by x, or 5 ft.

Represent the vert. dist. from base of tree to top of ladder by y, which is unknown.


Then (13 ft)^2 = (5 ft)^2 + y^2, or


169 ft^2 = 25 ft^2 + y^2. This simplifies to y^2 = 144. Thus y = + 12 feeet.


Note: Please pay attention to your spelling: "lader i up agenst a tree" should be "the top of a 13-ft ladder is placed against a tree."

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A firm’s marketing manager believes that total sales for next year will follow the normal distribution, with a mean of $3.2 mill
klasskru [66]

Answer:

The sales level that has only a 3% chance of being exceeded next year is $3.67 million.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

In millions of dollars,

\mu = 3.2, \sigma = 0.25

Determine the sales level that has only a 3% chance of being exceeded next year.

This is the 100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So X when Z = 1.88.

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

1.88 = \frac{X - 3.2}{0.25}

X - 3.2 = 0.25*1.88

X = 3.67

The sales level that has only a 3% chance of being exceeded next year is $3.67 million.

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