All categories

3 years ago

Solve for b. (b-7)m=r

Mathematics

1 answer:

deff fn [24]3 years ago

3 0

(b-7)m=r

Use the distributive property to multiply b-7 by m.

bm-7m=r

Add 7m to both sides.

bm=r+7m

The equation is in standard form.

mb=7m+r

Divide both sides by m.

mb/m = 7m+r/m

Dividing by m undoes the multiplication by m.

b= 7m+r/m

Divide r+7m by m.

b= r/m+7

You might be interested in

Question 5

zhannawk [14.2K]

Answer:

Your correct answer is B. If the price of the television was 31/5. the price of the laptop was $532

Step-by-step explanation:

You can divide it on a calculatoror on paper and get it correct! Hope it helped! Please mark brainliest! Have a wonderful day!!!

8 0

3 years ago

A teacher randomly selects one student from a group of students. 12 students are wearing a black shirt, 8 are wearing a red shir

Blababa [14]

12+8+8=28
8 students are wearing a blue shirt so the probability is 8/28

7 0

3 years ago

Read 2 more answers

Maria is planning to attend UCLA. She is curious what the average age of UCLA students is. Since most students that attend UCLA

RoseWind [281]

Answer:

Yes

We fail to reject the alternative hypothesis Hₐ < 30, that is the average age of students is less than 30

Step-by-step explanation:

Yes because, there is an average age of a sample which can be tested against a null hypotheses

We put

Null hypothesis as H₀ = 30

The alternative hypothesis as Hₐ < 30

We have

$z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

With

$\bar x$ = 29

μ = 30

σ = 5.2

n = 65

α = 10%

We have

Here we have z = -1.55 and critical z = -1.28

Which gives a critical $\bar x$ of 29.83 with the probability P = 0.061 < 0.1 Hence we reject the null hypothesis as there is sufficient evidence to suggest that the average age is less than 30 years.

8 0

3 years ago

What percent of 50 is 15

jarptica [38.1K]

Answer:

15 of 50 can be wriiten as:

$\frac{15}{50}$

Multiply both numerator & denominator by 100

$\frac{15}{50}$ × $\frac{100}{100}$ = $\frac{30}{100}$ = 30%

Hope this helps! Please mark brainliest. Have a nice day!

7 0

3 years ago

Read 2 more answers

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

Natasha_Volkova [10]

We have ∇f(x,y,z) = ⟨4x3,4y3,4z3⟩ and ∇g(x,y,z) = ⟨2x,2y,2z⟩, so LaGrange’s method gives requires that we solve the following system of equations:

x2 + y2 + z2 = 1
We split into four cases, depending on whether x and y are zero or not:

4x3 = 2λx 4y3 = 2λy 4z3 = 2λz

(1) (2) (3) (4)

(a) x and y are both nonzero. Then equations (1) and (2) tell us that x2 = y2 = λ/2, and putting √√ √√√

this into equations (3) and (4) gives solutions (± 2/2, ± 2/2, 0) and (± 3/3, ± 3/3, ± 3/3). (b) x̸=0buty=0. Thenwehavex2 =λ/2,from(1)andputtingthisinto(4)givesλ/2+z2 =1,

√√ which using (3) gives solutions (±1, 0, 0) and (± 2/2, 0, ± 2/2).

and (0, ± 2/2, ± 2/2).
(d) x = y = 0. Then equation (4) tells us that z = ±1, so we get the two solutions (0, 0, ±1).

Now we determine which of these points are maxima and minima by simply evaluating f at all these points. We find that the maximum value of f is 1 and occurs at (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1),

√√√√√√

while the minimal value is 1/3, and occurs at (± 3/3, ± 3/3, ± 3/3), (± 3/3, ± 3/3, ± 3/3), √√√

and (± 3/3, ± 3/3, ± 3/3).

4 0

3 years ago