What’s the slope?................

Which statement about 4x2 + 19x – 5 is true?

Thepotemich [5.8K]

Answer:

(x+5)

Step-by-step explanation:

$=>4x^2+19x-5\\=>4x^2+20x-x-5\\=>4x(x+5)-1(x+5)\\=>(x+5)(4x-1)$

Here, (x + 5) matches your answer choices

3 0

3 years ago

Read 2 more answers

Question a) Compare the Variability of Expenditure of Families in Two Towns given as follows: Number of families Expenditure (Ru

Karolina [17]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Number of families __________Expenditure

______midpoint (x) ____Town A __Town B

21-30 __25.5____________3 _______2

31-40 __35.5____________61 ______14

41-50 __45.5___________132 _____ 20

51-60 __55.5__________ 153 ______27

61-70 __65.5__________ 140 ______ 28

71-80 __75.5___________ 51 _______ 7

81-90 __85.5___________ 2 _______ 2

The variability is the variance and standard deviation :

The variance (s):

√Σ(x - m)²/Σf(x)

Town A:

m = Σ(X * f(x)) / Σf ; Σf = 542

Σ x * f(x) = (25.5*3) + (35.5*61) + (45.5*132) + (55.5*153) + (65.5*140) + (75.5*51) + (85.5*2)

= 29931 / 542

= 55. 22

Σf(x - m)² / Σf - 1 = [3(25.5-55.22)^2 + 61(35.5-55.22)^2 + 132(45.5-55.22)^2 + 153(55.5-55.22)^2 + 140(65.5-55.22)^2 + 51(75.5-55.22)^2 + 2(85.5-55.22)^2] / 541

= 76458. 4928 / 541

Variance = 141.32808

Standard deviation = √variance

Standard deviation = √141.32808

Standard deviation = 11.89

TOWN B:

m = Σ(X * f(x)) / Σf ; Σf = 100

Σ x * f(x) = (25.5*2) + (35.5*14) + (45.5*20) + (55.5*27) + (65.5*28) + (75.5*7) + (85.5*2)

= 5490 / 100

= 54.90

Variance :

Σf(x - m)² / Σf - 1 = [2(25.5-54.90)^2 + 14(35.5-54.90)^2 + 20(45.5-54.90)^2 + 27(55.5-54.90)^2 + 28(65.5-54.90)^2 + 7(75.5-54.90)^2 + 2(85.5-54.90)^2] / 99

= 16764 / 99

= 169.33

Standard deviation = √variance

Standard deviation = √169.3333

Standard deviation = 13.013

From the result above, Town A has lesser variability than Town B due to the slightly lower variance and standard deviation values.

8 0

3 years ago

Which figure has reflection symmetry?

seraphim [82]

The bottom left figure

5 0

2 years ago

Read 2 more answers

A computer programming team has 13 members. a. How many ways can a group of seven be chosen to work on a project? b. Suppose sev

Julli [10]

Answer:

1716 ;

700 ;

1715 ;

658 ;

1254 ;

792

Step-by-step explanation:

Given that :

Number of members (n) = 13

a. How many ways can a group of seven be chosen to work on a project?

13C7:

Recall :

nCr = n! ÷ (n-r)! r!

13C7 = 13! ÷ (13 - 7)!7!

= 13! ÷ 6! 7!

(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)

1235520 / 720

= 1716

b. Suppose seven team members are women and six are men.

Men = 6 ; women = 7

(i) How many groups of seven can be chosen that contain four women and three men?

(7C4) * (6C3)

Using calculator :

7C4 = 35

6C3 = 20

(35 * 20) = 700

(ii) How many groups of seven can be chosen that contain at least one man?

13C7 - 7C7

7C7 = only women

13C7 = 1716

7C7 = 1

1716 - 1 = 1715

(iii) How many groups of seven can be chosen that contain at most three women?

(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)

Using calculator :

(15 * 35) + (6 * 21) + (1 * 7)

525 + 126 + 7

= 658

c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?

(First in second out) + (second in first out) + (both out)

13 - 2 = 11

11C6 + 11C6 + 11C7

Using calculator :

462 + 462 + 330

= 1254

d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

Number of ways with both in the group = 11C5

Number of ways with both out of the group = 11C7

11C5 + 11C7

462 + 330

= 792

8 0

3 years ago

Half of a set of the parts are manufactured by machine A and half by machine B. Ten percent of all the parts are defective. Six

baherus [9]

Answer:

The probability that a part was manufactured on machine A is 0.3

Step-by-step explanation:

Consider the provided information.

It is given that Half of a set of parts are manufactured by machine A and half by machine B.

P(A)=0.5

Let d represents the probability that part is defective.

Ten percent of all the parts are defective.

P(d) = 0.10

Six percent of the parts manufactured on machine A are defective.

P(d|A)=0.06

Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective :

$P(A|d) =\frac{P(A \cap d)}{P(d)}$

$P(A|d) =\frac{P(d|A)\times P(A)}{P(d)}\\P(A|d)= \frac{0.06\times 0.5}{0.10}$

$P(A|d)= 0.3$

Hence, the probability that a part was manufactured on machine A is 0.3

8 0

3 years ago