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mixas84 [53]
2 years ago
15

Divide. 71 ÷ 6 Enter your answer by filling in the boxes. __R__

Mathematics
2 answers:
grin007 [14]2 years ago
5 0

Answer:

11R5

Step-by-step explanation:

To check answer 6×11=66 66+5=71

Andru [333]2 years ago
3 0
Possible answer: 11r8?
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Which statement about the equation 3/4y = 5/x is true?
VARVARA [1.3K]

Answer:

the answer would be no solution

Step-by-step explanation:

because, Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation

  3/4y-(5/x)=0

and from there you may not do anything else.

3 0
3 years ago
Randy sold corn for 12 ears for $2.How much would Randy make if he sold 600 ears of corn?
meriva

Answer:

$120

Step-by-step explanation:

12 ears = $2

600 ears = ?

to determine the cost of 600easrsof corn  we find the cost of 1 ears of corn first

1 ears = 2/12 =0.1666

= 0.2 approximately

so 600ears of corn = 0.2 × 600 =120

5 0
3 years ago
Helpppppp MEEEEE PLSSSS
saw5 [17]
18.84 all you do is the v times 3.14 and that’s what you get
7 0
3 years ago
Read 2 more answers
What is the value of p ?
VLD [36.1K]

Answer:

p = 35

Step-by-step explanation:

180 - 125 = 55

180 - 90 = 90

55 + 90 = 145

180 - 145 = 35

p = 35

6 0
3 years ago
The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

           \sigma   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

7 0
3 years ago
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