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3 years ago

No links plz and real answers

Mathematics

2 answers:

bearhunter [10]3 years ago

7 0

Answer:

Step-by-step explanation:

x³ + x²

4³ + 4²

64 + 16

romanna [79]3 years ago

5 0

Answer:

Step-by-step explanation:

$x^3+x^2$

$(4)^3+(4)^2$ Replace the variable x with 4 in the expression.

$64+(4)^2$ Raise 4 to the power of 3.

$64+16$ Raise 4 to the power of 2.

80 Add 64 and 16.

You might be interested in

The team won 5 out of 8 games if the team played 32 games how many did they win

Rus_ich [418]

Answer:

Step-by-step explanation:

32/8 = 4

4 x 5 = 20

7 0

3 years ago

In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore

erastovalidia [21]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

$p_{IS}$ true population proportion of high speed internet users that had been interrupted one or more times in the past month.

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

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

1) Part a

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:

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

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

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

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

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

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

If we replace the values obtained we got:

$0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316$

$0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524$

The 99% confidence interval would be given by (0.316;0.524)

3) Part c

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 $ME =\pm 0.05$ 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)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32$

And rounded up we have that n=335

4) Part d

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 $ME =\pm 0.05$ 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)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599$

And rounded up we have that n=649

5 0

3 years ago

7d^2+10 HELPPP SOVE THIS

Amanda [17]

So you just plug in the X value of the number that d represents for example if I plug in 2 for d

7(2)^2+10=38

4 0

3 years ago

How many different ways can the line be named? What are those names?

boyakko [2]

Answer: A. 3 ways: k, DE, ED (both DE and ED have line markers over top)

To name a line, we just need two points on the line. We list them in any order because the line extends forever in both directions. Contrast this with a ray where order does matter. The little k is another way to name a line, potentially simplifying things.

Choice B is close, but it mentions ray DE instead of line DE. Choice C is missing line ED. Choice D is a similar story as choice B. These facts allow us to rule out B through D.

5 0

3 years ago

Read 2 more answers

George buys some first class and second class stamps to put on some letters. Each first class stamp costs 67p, and each second c

Dafna1 [17]

Answer:

£191.36

Step-by-step explanation:

Sum the parts of the ratio, 1 + 4 = 5 parts

Divide 320 by 5 to find the value of one part of the ratio.

320 ÷ 5 = 64 ← value of 1 part of the ratio , thus

4 parts = 4 × 64 = 256

Thus he bought 64 first class and 256 second class.

Cost = (64 × £0.67) + (256 × £0.58)

= £42.88 + 148.48

= £191.36

3 0

3 years ago