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meriva
3 years ago
9

Which number line shows the correct placement of 20%, 1/3, and 2/3?

Mathematics
1 answer:
BigorU [14]3 years ago
7 0

Answer with Step-by-step explanation:

We are given that

20%

When we change percent into fraction we will divide the percent by 100

\frac{20}{100}=\frac{1}{5}

\frac{1}{3}

\frac{2}{3}

We have to find the line which shows the correct placement of given numbers.

Lcm(3,5)=15

\frac{1\times 3}{5\times 3}=\frac{3}{15}

\frac{1\times 5}{3\times 5}=\frac{5}{15}

\frac{2\times 5}{3\times 5}=\frac{10}{15}

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There were 3 bananas, 4 apples, and 3 oranges in a basket. What is the probability that Ace will pick a banana from the basket?
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0.3 or 30%. The probability that Ace pick a banana from a basket that content others fruits is 0.3.

The key to solve this problem is using the equation of probability P(A)=\frac{n(A)}{n} where n(A) the numbers of favorables outcomes and n the numbers of possible outcomes.

There are in the basket 10 fruits in total (3 bananas + 4 apples + 3 oranges = 10fruits). Then, extract a fruit can occur in 10 ways, this is n. There is only 3 bananas in the basket, so the fruit that ACE will pick be a banana can occur in 3 ways out of 10,  so 3 is n(A).

Solving the equation:

P(A)=\frac{3}{10}=0.3

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Answer: 12

Step-by-step explanation:

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In establishing warranties on HDTV sets, the manufacturer wants to set the limits so that few will need repair at the manufactur
Novosadov [1.4K]

Answer:

If the warranty limits are at 41.12 months, only 10 percent of the HDTVs need repairs at the manufacturer's expense.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. This means that \mu = 36.84, \sigma = 3.34.

Where should the warranty limits be set so that only 10 percent of the HDTVs need repairs at the manufacturer's expense?

This is the value of X when Z has a pvalue of 0.90.

Looking at the z-table, we get that this is between Z = 1.28 and Z = 1.29, so we use Z = 1.285.

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

1.28 = \frac{X - 36.84}{3.34}

X - 36.84 = 1.28*3.34

X = 41.12

If the warranty limits are at 41.12 months, only 10 percent of the HDTVs need repairs at the manufacturer's expense.

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Answer:

The product of y and 9 is less than 21

Step-by-step explanation:

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