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Ksju [112]
3 years ago
15

Write 0.2 as a fraction in simplest form.

Mathematics
2 answers:
stira [4]3 years ago
8 0

Answer:

1/5

Step-by-step explanation:

0.2 = 2/10

The reduced form of 2/10 is 1/5

Hence, the answer is 1/5

vredina [299]3 years ago
4 0

Answer:

1/5

Step-by-step explanation:

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Red bricks and grey bricks were used to build a decorative wall. The number of red bricks divided by number of grey bricks was 5
azamat
Red bricks : Grey bricks
        5        :            2

Total parts = 5 + 2 = 7

7 parts = 224
-----------------------
Find 1 part:
-----------------------
1 part = 234 ÷ 7
1 part = 32

-----------------------
Find 5 parts:
-----------------------
5 parts = 32 x 5 
5 parts = 160

----------------------------------------------
Answer: 160 Red bricks
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5 0
3 years ago
A public health organization reports that 40%of baby boys 6-8 months old in the United
Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

  • 0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
  • 0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
  • Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

P(X > 4) = 1 - P(X \leq 4)

In which:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Then:

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

P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061

P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403

P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150

P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502

Hence:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106

P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016

P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

4 0
1 year ago
A population of 360 field mice live inside of a small area of forest near your house.240 of the mice are brown, and 120 of the m
OlgaM077 [116]
It would be about 33%
4 0
3 years ago
A rock climber ascends 18 3/4 feet to the top of a rock ledge the climber descends 8 1/8
ratelena [41]

Answer:

10 10\frac{5}{8}

Step-by-step

Reduce \frac{3}{4} to \frac{6}{8} and simply subtract.

5 0
2 years ago
Write an equation for an ellipse centered at the origin, which has foci at (0,\pm\sqrt{63})(0,± 63 ​ )left parenthesis, 0, comma
steposvetlana [31]

Answer:

\frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }

Step-by-step explanation:

Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) \frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} }.

We find the co-vertices b from b = ±√(a² - c²) where a = 91 and c = 63

b = ±√(a² - c²)

= ±√(91² - 63²)

= ±√(8281 - 3969)

= ±√4312

= ±14√22

So the equation is

\frac{x^{2} }{(14\sqrt{22}) ^{2} } + \frac{y^{2} }{91^{2} } = \frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }

8 0
3 years ago
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