10. (07.07 LC)

The function f(x) = -(x+5)(x + 1) is shown

FinnZ [79.3K]

Answer:

all real numbers less than or equal to -4

Step-by-step explanation:

most of the numbers are negative less are positive

i need more information to help you better.

4 0

3 years ago

Solve rx = 30 for x.

yan [13]

X= 30/r is the answer because if you divide both sides by r you are left with x=30/r because you cant divide 30 by r

7 0

3 years ago

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$35 is 51% of what number?

babunello [35]

Answer:

-16

Step-by-step explanation:

$35 is 51% of what number is written as: $35 = 51% * (X)

$35 = 51% * (X)

35 - 51 = X

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<h2>Hope this helps!!!</h2>

6 0

3 years ago

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Please help!!

Morgarella [4.7K]

Answer:

Step-by-step explanation:(a)

1.65p+1.36j=118.17 and p+j=79

We can say p=79-j and sub this value in first equation

1.65(79-j)+1.36j=118.17

130.35-1.65j+1.36j=118.17

-0.29=-12.18

j=42, p=79-j, p=79-42=37

(b)

1.65(37)=61.05 and 1.36(42)=57.12

Sanchez class earned more selling pies.

(c)

61.05-57.12=3.93

So Sanchez class made $3.93 more than Kelley class.

7 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: