Answer true or false for the numbers 18 and 30. A common factor is 9. O True O False

A set of electric toothbrush prices are normally distributed with a mean of 87 dollars and a standard deviation of

Studentka2010 [4]

Answer:

The proportion of electric toothbrush prices are between 104.60 dollars and 108.20 dollars is 0.0099

Step-by-step explanation:

A set of electric toothbrush prices are normally distributed with a mean of 87 dollars and a standard deviation of 8 dollars.

$Mean = \mu = 87$

Standard deviation = $\sigma = 8$

We are supposed to find proportion of electric toothbrush prices are between 104.60 dollars and 108.20 dollars i.e.P(104.60<x<108.20)

$Z=\frac{x-\mu}{\sigma}$

At x = 104.60

$Z=\frac{104.60-87}{8}$

Z= 2.2

At x=108.20

$Z=\frac{108.20-87}{8}$

Z= 2.65

Refer the z table for p value :

P(x<108.20)-P(x<104.60)=P(Z<2.65)-P(Z<2.2)=0.9960-0.9861=0.0099

Hence The proportion of electric toothbrush prices are between 104.60 dollars and 108.20 dollars is 0.0099

4 0

3 years ago

Richard’s checking account balance was $57.34 at the beginning of the week. During the week, he recorded the transactions below.

Tomtit [17]

Answer:

US$ 132.45

Step-by-step explanation:

See attachment for the missing table.

Given:

Richard’s checking account balance at the beginning of the week = $57.34

Richard’s account balance at the end of the week from the given table:

Deposits of the week = US$ 163.75

Expenses of the week = Groceries + Credit card bill + Gas

Expenses of the week = 25.37 + 50 + 13.27

Expenses of the week = US$ 88.64

Richard’s account balance at the end of the week = Richard’s checking account balance at the beginning of the week + Deposits of the week - Expenses of the week

Replacing with the real values:

Richard’s account balance at the end of the week = 57.34 + 163.75 - 88.64

=US$ 132.4

7 0

3 years ago

Read 2 more answers

What are the amplitude, period, and phase shift of the given function? f(t)=-1/2sin(3t-2pi)

Romashka [77]

Answer:

The $amplitude=|A|=\frac{1}{2}$ , $period=3$ and $\text{phase shift}=\frac{2\pi}{3}$ .

Step-by-step explanation:

The given function is

$f(t)=-\frac{1}{2}\sin (3t-2\pi)$ .... (1)

The general form of sine function is

$F(t)=A\sin (Bt-C)+D$ ....(2)

where, |A| is the amplitude, B is period, D is the vertical shift (up or down), and C/B is used to find the phase shift.

On comparing (1) and (2), we get

$A=-\frac{1}{2}$

$B=3$

$C=2\pi$

$D=0$

So,

$amplitude=|A|=\frac{1}{2}$

$period=3$

$\text{phase shift}=\frac{2\pi}{3}$

Therefore $amplitude=|A|=\frac{1}{2}$ , $period=3$ and $\text{phase shift}=\frac{2\pi}{3}$ .

8 0

3 years ago

Factor 30b2 + 48b - 24

iogann1982 [59]

IM going to assume that the "2" is to be squared. If so, then the answer is 2(15x^2+8)

6 0

2 years ago

Use this graph of velocity vs. time for two objects to answer the question.

Lynna [10]

Using derivatives, it is found that the correct option is:

A. Object C has an acceleration that is greater than the acceleration for D.

The acceleration is the derivative of the velocity, given by change in velocity divided by change in time, that is:

$a = \frac{\Delta_v}{\Delta_t}$

In this problem:

The change in time for objects C and T is the same.
The change in velocity for object C is greater, thus, it has a greater acceleration, and the correct option is:

A. Object C has an acceleration that is greater than the acceleration for D.

A similar problem is given at brainly.com/question/14516604

6 0

2 years ago

Read 2 more answers