(a) Find the greatest common factor (GCF) of

Help help help help math

lidiya [134]

Answer:

Step-by-step explanation:

2x+y

=2(2)+1

=4+1

=5

3 0

2 years ago

Read 2 more answers

The board of examiners that administers the real estate broker's examination in a certain state found that the mean score on the

Goshia [24]

Answer:

The passing score is 645.2

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 553, \sigma = 72$

If the board wants to set the passing score so that only the best 10% of all applicants pass, what is the passing score?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28.

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

$1.28 = \frac{X - 553}{72}$

$X - 553 = 1.28*72$

$X = 645.2$

The passing score is 645.2

6 0

2 years ago

A cylindrical beaker with diameter 10 in and height 12 in. is filled with water that is then poured into a rectangular

emmasim [6.3K]

Answer:

A. The volume of the cylindrical beaker = 942.5 in3.

B. Volume of the pan = 378 in3

C. The water in the cylindrical beaker will over flow the pan.

D. The height of the water in the beaker after the pan is filled is 7.2 in

Step-by-step explanation:

Data given from the question.

Diameter = 10 in

Height = 12 in

Dimension of the pan = 14 in x 9 in x 3 in

Determination of the volume of each solid.

A. Volume of the cylindrical beaker:

Volume = πr2h

Radius (r) = 10/2 = 5 in

Height (h) = 12 in

Volume (V) =?

V = πr2h

V = π x (5)^2 x 12

V = 942.5 in3.

The volume of the cylindrical beaker = 942.5 in3.

B. Volume of the rectangular pan

Volume = Length x Width x Height

The dimension for the rectangular pan = volume of the pan = 14 in x 9 in x 3 in

Volume of the pan = 378 in3

C. Since the volume of the cylindrical beaker is higher than that of the pan, the water in the cylindrical beaker will overflow the pan.

D. Determination of the height of the water in the cylindrical beaker after the pan is filled. This is illustrated below.

Let us calculate the volume of the water in the cylindrical beaker after the pan is filled

The volume of the cylindrical beaker = 942.5 in3.

Volume of the pan = 378 in3

Volume of water in the cylinder beaker after the pan is filled = 942.5 - 378 = 564.5 in3

With this new volume, we can calculate the new height of the water as follow:

Volume = 564.5 in3

Radius (r) = 5 in (the radius remains the same)

Height (h) =?

V = πr2h

h = V/πr2

h = 564.5/π(5)^2

h = 7.2 in

Therefore, the height of the water in the beaker after the pan is filled is 7.2 in

4 0

2 years ago

Use the Quadratic Formula to find the exact solutions of x2 + 7x - 4 = 0.

sladkih [1.3K]

Answer:

The solution of equation $x^2+7x-4=0$ is 0.53,-7.53.

Step-by-step explanation:

Given : Quadratic equation $x^2+7x-4=0$

To find : The solution of equation ?

Solution :

The solution of the quadratic equation $ax^2+bx+c=0$ is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=1, b=7 and c=-4.

Substitute the values,

$x=\frac{-7\pm\sqrt{7^2-4(1)(-4)}}{2(1)}$

$x=\frac{-7\pm\sqrt{65}}{2}$

$x=\frac{-7+\sqrt{65}}{2},\frac{-7-\sqrt{65}}{2}$

$x=0.53,-7.53$

Therefore, the solution of equation $x^2+7x-4=0$ is 0.53,-7.53.

3 0

2 years ago

What is the value pqr/pq-qr if p=-3,q=2 and r=2?

yaroslaw [1]

Answer:

1.2 OR 1 1/5

Step-by-step explanation:

-3 x 2 x 2 = -12

-3 x 2 = -6

2 x = 4

-12 / -6 - 4 =

-12 / -10 =

1.2 OR 1 1/5

Hope that helps!

6 0

3 years ago