Please i really need help

Can anyone help me solve this?

xxMikexx [17]

Answer:

a- 16/100

b- 1/6

c- not sure

Step-by-step explanation:

8 0

4 years ago

Evaluate (if possible) the six trigonometric functions of the real number. (If not possible, enter IMPOSSIBLE.)

kotegsom [21]

$\sin(t) = \frac{1}{2} 
\\
\\ \cos(t) = - \frac{ \sqrt{3} }{2} 
\\
\\ \tan(t) = \frac{\sin(t)}{\cos(t)}= - \frac{1}{ \sqrt{3} } 
\\
\\ \csc(t) = \frac{1}{\sin(t)}= 2 
\\
\\ \sec(t) =\frac{1}{\cos(t)}=- \frac{2}{ \sqrt{3}} 
\\
\\ \cot(t) = \frac{1}{\tan(t)}=- \sqrt{3}$

3 0

3 years ago

Rita has 3/4 gallon of lemonade. She pours 1/8 gallon of lemonade into each glass. How many glasses can she pour?

bearhunter [10]

Answer: 6 glasses.

Step-by-step explanation:

You have to multiply the denominator of 3/4 by two and the numerator by 2 to get 6/8. If you're pouring 1/8 into each glass, then you should be able to have 6.

8 0

3 years ago

Determine whether each statement is sometimes, always, or never true. Two solids with equal volumes are congruent. Select one: a

navik [9.2K]

A. sometimes
just because they have the same volume does not mean they are congruent. they could have different heights and base areas
hope this helps:)

3 0

4 years ago

Read 2 more answers

A manufacturer of small appliances purchases plastic handles for coffeepots from an outside vendor. If a handle is cracked, it i

Rus_ich [418]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04$

And rounded up we got:

$n\approx 97$

Step-by-step explanation:

Data given and previous concepts

ME=0.1 represent the margin of error desired

Confidence =0.95 or 95%

$\alpha=0.05$ represent the significance level

z represent the quantile from the normal standard distribution

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Since we don't have a prior estimate for the proportion $\hat p$ we can use as estimator 0.5 and replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04$

And rounded up we have that n=97

5 0

4 years ago