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2 years ago

12

55% of the attending crowd was female. If the crowd has 1345 people, how many females are there ?

1 answer:

Assoli18 [71]2 years ago

3 0

Not sure but maybe 740

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7% of 150 = 10.5

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What could be the dimensions of a triangle with an area of 12in square?

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The area is about 62.4 inch square

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Will make you brainlistt if it is right

dexar [7]

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The amount of weeks is your independent variable, or X variable, and the height is you dependents variable.

Step-by-step explanation:

I don't know what this question is about so I'm gonna use plants as an example. The amounts of weeks that you are growing a plant is not dependent on anything. Nothing can cause change to that variable, so it would be your X. The height that your plant has grown DOES depend on how long you've grown it, so the amount of weeks, making it your Y variable.

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3 years ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

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

$\alpha = \frac{1 - 0.98}{2} = 0.01$

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 $1 - 0.01 = 0.99$ , so Z = 2.327.

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.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

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

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

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2 years ago

6. I think the answer is H, could you help and check?!

Sliva [168]

Set up an equation to solve.
70% of original price equals $550, the amount Susan is paying.
0.70x = 550

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3 years ago

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