All categories

3 years ago

What is the answer to 7b-b=-6

2 answers:

8 0

Answer: B= -1
•••Simplify both sides of the problem. 7b−b=−6
•••Now we can simplify step by step.
7b−b=−6
7b+−b=−6
(7b+−b)=−6
6b=−6
•••Divide both sides by 6.
6b/6=−6/6 and that equels -1.
So b= -1
Hope this helps!

sergejj [24]3 years ago

3 0

6b=-6
b=-1
my answer need to be long, that is why I write this sentence

You might be interested in

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

Draw a line plot to correctly display the data 4,5,6,9,13,14,14,14,15 mean= median= mode= range=

ziro4ka [17]

Mean=10.4 since you add all the numbers which gives you 94 and divide by how many they are and that is 9
median=13 (middle number)
mode=14 constant seen number
range= 11 because 15 minus 4 is 11 (difference between your lowest and highest value
a line plot is another word for dot plot so just use 1 -20 and arrange the number
hope this helps

5 0

3 years ago

PLZ HELP ITS DUE T MIDNIGHT

kkurt [141]

I don’t know the answer but I can tell you how to do it subtract starting value minus final value divide that amount by the absolute value of the starting value multiply by 100 to get percent decrease if the percentage is negative it means there was an increase not a decrease

5 0

2 years ago

(5x ² +1) (2x + 3) - 2 x (5²-3)

kotegsom [21]

Answer:

y = 3 suh slid este ob cine este a, b și g au nevoie de acel kinfo pentru a termina

Step-by-step explanation:

6 0

3 years ago

The equation 2x^2 + x - 1 = 0 has two solutions. Find an equation of the form ax^2 + bx + c = 0, which solutions....

leonid [27]

Answer:

Step-by-step explanation:

Let the solution to

2x^2 + x -1 =0

x^2+ (1/2)x -(1/2)

are a and b

Hence a + b = -(1/2) ( minus the coefficient of x )

ab = -1/2 (the constant)

A. We want to have an equation where the roots are a +5 and b+5.

Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.

The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.

So the equation is

x^2-(19/2)x + 22 =0

2x^2-19x + 44 =0

B. We want the roots to be 3a and 3b.

Hence (3a) + (3b) = 3(a+b) = 3(-1/2) =-3/2 and

(3a)(3b) = 9(ab) =9(-1/2)=-9/2.

So the equation is

x^2 +(3/2) x -9/2 = 0

2x^2 + 3x -9 =0.

4 0

3 years ago