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3 years ago

What is the approximate perimeter of a rhombus with the diagonals that measure 12 feet and 18 feet

Mathematics

1 answer:

Fynjy0 [20]3 years ago

7 0

Answer:

The perimeter of a rhombus with diagonals of 12 feet and 18 feet would be 60 feet.

Step-by-step explanation:

Perimeter of a rhombus: S1 + S2 + S3 + S4 = answer. (S stands for side.)

12 ft. + 18 ft. + 12 ft. + 18 ft. = 60 ft.

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PLZ HELP ILL MARK U BRAINLIEST

Dafna11 [192]

30 b/c 60 x 1/2= 30

Answer= 30

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3 years ago

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Find the prime factorization on 168

Paha777 [63]

Answer:

The prime factors of 168 are 2, 3, and 7.

Step-by-step explanation:

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2 years ago

3) Two of these numbers are the same value, which ones are<br> they? 40%, .44, 14, .04, 4/10?

baherus [9]

Answer:

40% and 4/10

Step-by-step explanation:

If you convert them into decimals, they will both give the same decimals which gives 0.4

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2 years ago

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who does the number, of zeros in the product of 8 and 5000 compara to the number of zeros in the factors ? explain.

mezya [45]

If one of the numbers we multiply (factors) has zeros at the end, and the other isn't a fraction: all those zeros will stay in the product.

But there might be additional zeros if the other numbers in the factors (the numbers which aren't 0) mupliply to "end" in zero and this is the case here:

8*5=40.

so the product will be 40 and the zeros of the 5000:

40 000

the number of zeros in the product will be bigger than the number of zeros in the factors if the non-zero parts of the fractions multiply to a number with 0 at the end.

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2 years ago

A cookie factory monitored the number of broken cookies per pack yesterday.

trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

$s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }$ $\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }$

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

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2 years ago