Algebra 1, Section 7.09

a mechanical engineer designs a robotic rm to be used in an assembly line. one portion of the arm is 5/12 yard long, and the len

Amanda [17]

Answer:

8/12 yard or 2/3 yard

Step-by-step explanation:

The sum of the two lengths is ...

5/12 yd + 3/12 yd = (5+3)/12 yd = 8/12 yd

This fraction can be reduced by removing a factor of 4 from numerator and denominator.

8/12 yd = 2/3 yd

If the arm portions are end-to-end, the total length is 2/3 yard.

4 0

3 years ago

The probability that a 38-year-old white male will live another year is .99813. What premium would an insurance company charge t

alukav5142 [94]

Answer:

The insurance company should charge $1,873.5.

Step-by-step explanation:

Expected earnings:

1 - 0.99813 = 0.00187 probability of the company losing $1 million(if the client dies).

0.99813 probability of the company earning x(price of the insurance).

What premium would an insurance company charge to break even on a one-year $1 million term life insurance policy?

Break even means that the earnings are 0, so:

$0.99813x - 0.00187(1000000) = 0$

$0.99813x = 0.00187(1000000)$

$x = \frac{0.00187(1000000)}{0.99813}$

$x = 1873.5$

The insurance company should charge $1,873.5.

4 0

2 years ago

What is the sum of product of a number x and 14 , and 10

malfutka [58]

14x+10 asldfkjalsdkf

7 0

3 years ago

:

Alinara [238K]

There was 4.35% error in Diego's measurements.

Step-by-step explanation:

Given,

Length measured by Diego = Approx value = 22 cm

Actual length = Exact value = 23 cm

Percent error = $\frac{|approx-exact|}{exact}*100$

$Percent\ error = \frac{|22-23|}{23}*100\\\\Percent\ error = \frac{|-1|}{23}*100\\\\Percent\ error = \frac{1}{23}*100\\\\Percent\ error = \frac{100}{23}\\\\Percent\ error = 4.347\%$

Rounding off to nearest hundredth

Percent error = 4.35%

There was 4.35% error in Diego's measurements.

Keywords: percentage, subtraction

Learn more about percentages at:

#LearnwithBrainly

5 0

2 years ago

Suppose that for a particular hypothesis test, the consequences of a Type I error are not very serious, but there are serious co

VARVARA [1.3K]

Answer:

We want to reduce type II error we carry out the test using a larger significance level (such as 0.10) and not a small significance level α (such as 0.01).

Step-by-step explanation:

Type I error

Rejecting the null hypothesis when it is in fact true is called a Type I error.
It is denoted by alpha, α that is the significance level.
Lower values of alpha make it harder to reject the null hypothesis, so choosing lower values for alpha can reduce the probability of a Type I error.

It is given that the consequences of a Type I error are not very serious, but there are serious consequences associated with making a Type II error.

Type II error

This is the error when we fail to reject a false null hypothesis or accept a null hypothesis when it is true.
Higher values of alpha makes it easier to reject the null hypothesis.
So choosing higher values for alpha can reduce the probability of a type II error.
The consequence here is that if the null hypothesis is true, increasing the value of alpha makes it more likely that we make a Type I error.

Since, we want to reduce type II error we carry out the test using a larger significance level (such as 0.10) and not a small significance level α (such as 0.01).

This will increase type I error but that is okay since we do not have serious consequences for it.

5 0

3 years ago