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3 years ago

What is the approximate measure of the leg opposite of angle A???

Mathematics

1 answer:

creativ13 [48]3 years ago

4 0

Answer:

8.7 cm

Step-by-step explanation:

take 60 degree as reference angle

using sin rule

sin 60=opposite/hypotenuse

$\frac{\sqrt{3} }{2}$ =opposite/10

$\frac{\sqrt{3} }{2} * 10$ =opposite

$\sqrt{3} *5$ = opposite

$5\sqrt{3}$ =opposite

8.66=opposite

8.7 =opposite

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The area to the right of z = -0.60, rounded to four decimal places, is

TEA [102]

You need to a table of the standard normal cumulative distribution
Here is one:
https://math.ucalgary.ca/files/math/normal_cdf.pdf

0.2743 is what I see

5 0

4 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.

7 0

3 years ago

A factory fills bottles with a beverage, and each bottle is supposed to contain 500\text{ mL}500 mL500, start text, space, m, L,

Delicious77 [7]

Answer:

The test statistic for Norah's test is $t = -2$

Step-by-step explanation:

Norah is in charge of a quality control test that involves measuring the amounts in a sample of bottles to see if the sample mean amount is significantly different than 500 ml.

This means that at the null hypothesis we test if the sample mean is 500 ml, that is:

$H_0: \mu = 500$

At the alternate hypothesis, we test if it is differente than 500 ml, that is:

$H_a: \mu \neq 500$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.