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2 years ago

Decimal formal 1.96 becomes what expressed as a percent

Mathematics

1 answer:

olasank [31]2 years ago

7 0

Expressed as a percent, it becomes 196%

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BADEHGJK NM because it's going backwards and then it skips a letter

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What does 9y =27 equation mean

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9y=27

divide by 9 each side

9y/9y=27/9

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Two buses are moving towards each other, one at a speed of 40 mph and the other at a speed of 50 mph. How much closer to each ot

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A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their pur

Harrizon [31]

Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.8}{2} = 0.1$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.1 = 0.9$ , so Z = 1.28.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.28\frac{1.5}{\sqrt{305}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

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3 years ago

Consider an urn containing 8 white balls, 7 red balls and 5 black balls.

weqwewe [10]

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

$=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053$

2) the probability of getting 2 white balls is equal to:

$=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465$

3) The probability of getting at least 72 white balls is:

$=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72} \times \left( \frac{7}{20} \right)^{78} +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73} \times \left( \frac{7}{20} \right)^{77} + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149} \times \left( \frac{7}{20} \right)^{1} +\left( \frac{8}{20} \right)^{150}$

$=\sum^{150}_{k=72} [C^{k}_{150}\times \left( \frac{8}{15} \right)^{k} \times \left( \frac{7}{15} \right)^{150-k}]$

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10 months ago

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