The number of blogs has grown rapidly. Assuming that two new blogs are created each second,

Nesterboy [21]

51/6(n - 4) + 51/6 n = 442

51/6 n - 51/6(4) + 51/6 n = 442
17n - 34 = 442
17n = 408
n = 27

answer
Dustin sold 27 packs

4 0

3 years ago

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Find the missing side

saul85 [17]

Answer:

x = 14

$h^2 = p^2 + b^2\\h^2 = 7^2 + 12^2\\x^2 = 49 + 144\\x^2 = 193\\so, x = 13.89\\so, x = 14$

6 0

3 years ago

I need help on question 2 i need the simplified answer and the restrictions

Anastaziya [24]

Answer:

x cannot be 5 or -3.

$\frac{x+5}{x+3}$

Step-by-step explanation:

The restrictions for a fraction is that the bottom cannot be 0.

So if we find when the bottom is 0 we have found the values that x cannot be.

Let's solve x^2-2x-15=0.

Since the coefficient of x^2 is 1 all we have to do is find two numbers whose product is -15 and whose sum is -2.

Those numbers are -5 and 3 since (-5)(3)=-15 and (-5)+(3)=-2.

So the factored form of the equation is:

(x-5)(x+3)=0

This means either x-5=0 or x+3=0.

We do have to solve both.

x-5=0 can be solved by adding 5 on both sides.

x-5+5=0+5

x+0=0+5

x=5

x+3=0 can be solved by subtracting 3 on both sides.

x+3-3=0-3

x+0=0-3

x=-3

So x can be any number except x=5 or x=-3.

We already factored the bottom as (x-5)(x+3).

The top is a difference of squares, x^2-a^2,

which can be factored as (x-a)(x+a).

So the top factors as (x-5)(x+5).

The fraction can then be written as:

$\frac{x^2-25}{x^2-2x-15}=\frac{(x-5)(x+5)}{(x-5)(x+3)}$

This can further simplified assuming x is not 5 we can write it as $\frac{x+5}{x+3}$ . I canceled the common factor of (x-5).

5 0

3 years ago

Can someone help me with this?

hodyreva [135]

Use inverse to solve this problem. Tan-1(5/27). =10.491 which rounded to the nearest degree equals 11.

8 0

3 years ago

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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: