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2 years ago

How to solve scientific notation numbers

1 answer:

4 0

Look at the exponent and move the decimal that many places ( to the right for a positive number and to the left for a negative number)

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Of 200 people surveyed, 80 percent own a cellular phone and 45 percent own a pager. If all 200 people surveyed own a cellular ph

Ksju [112]

Answer:Of 200 people surveyed, 80 percent own a cellular phone and 45 ... If all 200 people surveyed own a cellular phone, or a pager, or both, .... Percent of those surveyed either do not own a cellular phone = 80% - 25% = 55%

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3 years ago

A rectangle has a side length of 3 feet and an area of 24 squared feet. A larger similar triangle has a corresponding length of

Zinaida [17]

3*w=24
w=8
9/3=3
8*3=24

9*24=216

7 0

3 years ago

Aleksandr [31]

Answer: lower bound = 9550

<u>Step-by-step explanation:</u>

Since the number is rounded to the nearest hundred, the actual value is somewhere between 9550 (which rounds up to 9600) and 9649 (which rounds down to 9600).

The lower bound is: 9550

The upper bound is: 9649

6 0

2 years ago

Mark drank 3 2/5 cups of orange jucie in the morning and 2 1/5 cups in the afternoon. How much orange jucie did mark drink in al

Alinara [238K]

Solutions

1) Convert mixed fractions into improper fractions.

$\frac{17}{5}+ \frac{11}{5}$

$= ((17 * 5) + (11 * 5)) / (5 * 5)$

$= (85 + 55) / 25$

$= 140/25$

$= 28/5$

$5 \frac{3}{5}$

5 0

2 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

2 years ago