All categories

3 years ago

What number goes in the box to make a true statement?

Mathematics

1 answer:

LenaWriter [7]3 years ago

5 0

Answer:

Sperm is the male reproductive cell, or gamete, in anisogamous forms of sexual reproduction. Animals produce motile sperm with a tail known as a flagellum, which are known as spermatozoa, while some red algae and fungi produce non-motile sperm cells, known as spermatia

You might be interested in

Need help with both of these please!! ASAP

Alik [6]

x = 1

Answer:

$m\angle XVW = 100\degree$

Step-by-step explanation:

In the first figure, FP is the bisector of $\angle DFE$

$\therefore m\angle 1 = m\angle 2\\\therefore 44x + 1 = 46x - 1\\\therefore 1 + 1 = 46x - 44x\\\therefore 2 = 2x\\\therefore \frac{2}{2} = x \\\huge \red {\boxed {\therefore x = 1}}$

In the second figure, VP is the bisector of $\angle XVW$

$\therefore m\angle 1 = m\angle 2\\\therefore 12x + 2 = 11x +6\\\therefore 12x - 11x = 6 - 2\\\therefore x = 4\\\because m\angle XVW = m\angle1 + m\angle 2\\\therefore m\angle XVW = 12x + 2 + 11x +6\\\therefore m\angle XVW = 23x + 8\\\therefore m\angle XVW = 23\times 4+ 8\\\therefore m\angle XVW = 92+ 8\\\huge \purple {\boxed {\therefore m\angle XVW = 100\degree}} \\$

8 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$