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salantis [7]
2 years ago
13

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

Mathematics
1 answer:
Mkey [24]2 years ago
5 0
The Answer is C, 5.27 x 10^6

Please mark it as brainlest answer:).

You might be interested in
Please Help!! question is attstched!
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
Read 2 more answers
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
Read 2 more answers
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
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