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Rainbow [258]
3 years ago
8

If 4 rocks fit in a box how many boxes would it take to hold 32 rocks

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
8 0

Answer:

it would take 8 boxes

Step-by-step explanation:

thirty two divided by four is eight

Orlov [11]3 years ago
6 0

Answer:

8 boxes. 32÷4 = 8

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Me pueden ayudar a saber el resultado y la operacion de la ecuacion 8-5x=8+2x por favor me urge
Art [367]
Solve for x:
8 - 5 x = 2 x + 8
Subtract 2 x from both sides:
8 + (-5 x - 2 x) = (2 x - 2 x) + 8
-5 x - 2 x = -7 x:
-7 x + 8 = (2 x - 2 x) + 8
2 x - 2 x = 0:
8 - 7 x = 8
Subtract 8 from both sides:
(8 - 8) - 7 x = 8 - 8
8 - 8 = 0:
-7 x = 8 - 8
8 - 8 = 0:
-7 x = 0
Divide both sides of -7 x = 0 by -7:
(-7 x)/(-7) = 0/(-7)
(-7)/(-7) = 1:
x = 0/(-7)
0/(-7) = 0:
Answer: x = 0
6 0
3 years ago
The graph of f(x) = –3(x – 1)2 + 4 is shown on the left. Which interval defines the domain of the function? (-∞, 4) (-∞, 4] (-∞,
noname [10]

Answer:

C

Step-by-step explanation:

The graph can take any value of X

-3[ x^2 - 2x + 1] + 4

-3x^2 +6x -3 + 4

-3x^2 + 6x + 1

x can take any value of real numbers and there would be a solution

Hence all real numbers is the domain.

6 0
3 years ago
Convert 6,250 pounds to tons
MariettaO [177]

Answer:

6250 pounds =3.125 us tons

8 0
4 years ago
Read 2 more answers
To play the lottery in a certain state, a person has to correctly select 5 out of 45 numbers, paying a $1 for each five-number s
defon

There are 45 numbers out of which 5 numbers are required to win an enormous sum of money

No of ways  in which 5 numbers can be selected out of 45 numbers.

C_{45}^5=\frac{45!}{40!5!} =1221759

out of which there is only combination foe contest winning

P=\frac{1}{1221759}

When 100 different tickets are bought then

Probability of winning: \frac{100}{1221759}

Probably of winning≈0,000082

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\

P.S. Hello from Russia :^)

4 0
3 years ago
A publisher reports that 344% of their readers own a particular make of car. A marketing executive wants to test the claim that
inysia [295]

Answer:

No, there is not enough evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.

And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;

Null Hypothesis, H_0 : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}

Alternate Hypothesis, H_1 : p \neq 0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}

The test statistics we will use here is;

                T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } } ~ N(0,1)

where, p = actual % of readers who own a particular make of car = 0.34

            \hat p = percentage of readers who own a particular make of car in a

                  sample of 220 = 0.30

            n = sample size = 220

So, Test statistics = \frac{0.30 -0.34}{\sqrt{\frac{0.30(1- 0.30)}{220} } }

                             = -1.30

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.

3 0
3 years ago
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