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3 years ago

Solve the equation. Show your work. 85 = 6b + 19

Mathematics

2 answers:

Elodia [21]3 years ago

8 0

This is how you do it:

85 = 6b + 19

85 - 19 = 6b + 19 - 19

66 = 6b

Divide 66 and 6b by 6

11 = b

patriot [66]3 years ago

8 0

Hope this helps! Message me with any questions :)

You might be interested in

List 3 values that would make this in quality true. 48<6n

Helen [10]

The 3 values would be 9 , 10 and 11.

because:

48<6(9)
48<54

48<6(10)
48<60

48<6(11)
48<66

6 0

3 years ago

Hailey wants to put a skirt around her desk to hide all her computer equipment and cords. She measured her desk, found the dimen

xeze [42]

Answer:

$C.\ Perimeter = 48 * 18$

Step-by-step explanation:

Given

See attachment for her sketch

Required

Which equation do not represent the perimeter

From the attached sketch:

$L =18in$ --- Length

$W =48in$ --- Width

Perimeter (P) is calculated as:

$P = 2 * (L + W)$

This gives:

$P = 2 * (18 + 48)$ ---- This represents (A)

Open bracket

$P = 2 * 18 + 2*48$ ---- This represents (B)

In algebra:

$2 * a$ means $a+ a$

So, the expression becomes

$P = 18 + 18 +48 + 48$ --- This represents (D)

<em>This implies that (C) does not represent the perimeter</em>

4 0

2 years ago

You work for a company that makes tiles for patios. A customer sent you the following picture of his patio. He said the patio is

Mumz [18]

Answer:3 by 1

Step-by-step explanation:

okay so I'll say the long side of a tile as x

the short side of a tile as y

look at the short side of the big patio first, it's made up of 2 big and 2 small

so 2x+2y=8 or x+y=8 so two added together is 8. So one possibility is 3 and 1 because (3+1)2 = 9

8 0

2 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

Deshawn has two bags of marbles. The first bag has 2 blue, 3 orange, and 5 red. The second bag has 4 pink, 10 blue, and 6 brown.

madreJ [45]

Answer: 0.1

Step-by-step explanation:

Given

Bag-I has 2 blue,3 orange, 5 red

Bag-II has 4 Pink,10 blue, 6 brown

No of ways of choosing a blue marble from bag-I

$\Rightarrow ^2C_1$

Total no of ways of choosing a marble from bag-I

$\Rightarrow ^{10}C_1$

No of ways of choosing a blue marble from bag-II

$\Rightarrow ^{10}C_1$

Total no of ways of choosing a marble from bag-II

$\Rightarrow ^{20}C_1$

The probability that he will pull out a blue marble from each bag is

$\Rightarrow P=\text{Probability of pulling a blue marble from bag-I}\times \text{Probability of pulling a blue out bag-II}$

$\Rightarrow P=\dfrac{^2C_1}{^{10}C_1}\times \dfrac{^{10}C_1}{^{20}C_1}\\\\\Rightarrow P=\dfrac{2}{10}\times \dfrac{10}{20}=\dfrac{1}{10}$

3 0

3 years ago