Where do erasers go for vacation? missing lengths

H (x) = -x+4 when h(x)=-3

Aleks04 [339]

X=7
. hope that helps

5 0

4 years ago

If you answer all these questions for me, you'll get extra points on brainly!

dmitriy555 [2]

Answer:

its too small :(

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the volume of the cone below?

lara [203]

Answer:

the world may never know

Step-by-step explanation:

7 0

3 years ago

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct a 95% conden

S_A_V [24]

Answer:

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

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 $1 - \frac{\alpha}{2}$ .

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead.

This means that $n = 42, \pi = \frac{26}{42} = 0.619$

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 $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 - 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.472$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 + 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.766$

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

4 0

3 years ago

A tire company has developed a new type of steel-belted radial tire. Extensive testing indicates the population of mileages obta

Andru [333]

Answer:

Step-by-step explanation:

According to the given question, a tire company has developed a new type of steel-belted radial tire. Extensive testing indicates the population of mileages obtained by all tires of this new type is normally distributed with a mean of 37,000 miles and a standard deviation of 3,887 miles.

Let us define X be the random variable shows that the mileages tires normally distributed with

mean

μ = 37000

standard deviation

σ =3, 887

Therefore

X ~ (μ = 37000, σ =3,887)

The company wishes to offer a guarantee providing a discount on a new set of tires if the original tires purchased do not exceed the mileage stated in the guarantee. Therefore the guaranteed mileage be if the tire company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage is determined as:

P(X < k) = 0.02

$\Rightarrow P\left ( \frac{X-\mu }{\sigma }\leq \frac{k-\mu }{\sigma } \right )=0.02
\\\\P\left ( Z \leq \frac{k-37,000 }{3,887 } \right )=0.02$

From the standard normal curve 2% area is determined as -2.0537 and hence

$\frac{k-37,000}{3,887}=-2.0537\\\\k=37000-7982.7319\\\\k=29017.2681\\\\\Rightarrow k=29018$

If we consider z value at two decimal places then

$\frac{k-37,000}{3,887} =-2.05\\\\\Rightarrow k=29032
$

Therefore the guaranteed 29032 mileage be if the tire company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage.

The area under the standard normal curve is determined as:

6 0

4 years ago