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Mila [183]
3 years ago
7

Help? anyone 8th grade math

Mathematics
2 answers:
Zinaida [17]3 years ago
8 0

Answer: x = -1

Step-by-step explanation:

Sveta_85 [38]3 years ago
4 0

Answer:

-1

Step-by-step explanation:

\frac{8.7+x}{1.1} = 7

8.7 + x = 7.7

x = -1

\therefore the value of x is -1

Hope this helps!

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Solve each equation algebraically. 2/3x( - 4)=20
baherus [9]

Answer:

Step-by-step explanation:

2/3x(-4)=20

2/3x*4=20

8/3x=20

8/3x/8/3= 20/8/3

0.83 but 3 is repeting

8 0
3 years ago
Strawberries cost $1.19 per pound. What is the cost of 2.16 pounds of strawberries?
maksim [4K]

Answer:

$ 2.57

Step-by-step explanation:

1.19 $/pound  * 2.16 pound =

7 0
2 years ago
I need assistance with this math problem.
Harman [31]
The answer is 80sq.ft... to figure this out, you just multiply both sides together
8 0
3 years ago
Solve for x, <br> 3(x – 4) ≥ 5x + 2
Flauer [41]

\Large\textsc{answer :}

x≤-7

\Large\textsc{calculations :}

First , use the distributive property :

3(x-4)≥5x+2

3x-12≥5x+2

Subtract 5x on both sides :

3x-5x-12≥2

Now , add 12 on both sides :

3x-5x≥2+12

3x-5x≥14

Subtract the x's :

-2x≥14

Divide by -2 on both sides and watch what happens :

x≤-7 (notice that the inequality sign changed from "greater than or equal to" to "less than or equal to")

If we divide both sides of an inequality by a negative number , we change the inequality sign .

\footnotesize\textit{hope helpful~}

7 0
2 years ago
67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r
oksian1 [2.3K]

Answer:

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p  = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

P(X=r) = nCr p^r q^{n-r}

where nCr = \frac{n!}{(n-r)! r!}

a) Probability that exactly 29 of them are spayed or neutered

P(X= 29) = 48C29 * 0.67^{29} * 0.23^{19}\\P(X=29) = 0.074

b) Probability that at most 33 of them are spayed or neutered

P(X \leq 33) =1 -  P(X > 33)\\P(X \leq 33) =1 - 0.34\\P(X \leq 33) = 0.66

c) Probability that at least 30 of them are spayed or neutered

P(X \geq 30) = 1 - P(x < 30)\\P(X \geq 30) = 1 - 0.21\\P(X \geq 30) = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

P(28 \leq X \leq 33) = P(X=28) + P(X=29) + P(X=30) + P(X=31) + P(X=32) + P(X=33)\\P(28 \leq X \leq 33) = 0.053 + 0.074 + 0.095 + 0.112 + 0.121 + 0.119\\P(28 \leq X \leq 33) = 0.574

8 0
3 years ago
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