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Aneli [31]
3 years ago
9

What is the least common multiple of 12 32

Mathematics
2 answers:
harina [27]3 years ago
7 0

Answer: the answer is 96


Step-by-step explanation:


iogann1982 [59]3 years ago
4 0

Step-by-step explanation:The answer is 2

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Help me out please .
alex41 [277]

The first one

Step-by-step explanation:

1/7 of 42 is 6

2/3 of 42 is 28

6+28=34

42-34=8

8/42=4/21

6 0
3 years ago
Help me out with this one please
Anna11 [10]

Answer:

5) x = -1

6) x = 2

Step-by-step explanation:

5) 5 + 2x + 6 = x + 10

   2x + 11 = x + 10

   2x + 1 = x

   1 = -x

   -1 = x

6) (9x + 7) + (-3x + 20) = 39

    6x + 27 = 39

    6x = 12

     x = 2

8 0
3 years ago
Read 2 more answers
A swimming pool can be filled by any of three hoses A, B or C. Hoses A and B together take 4 hours to fill the pool. Hoses A and
yanalaym [24]

Answer:

3 hours  28 minutes

Step-by-step explanation:

Let A = the RATE at which hose A can fill the pool alone

Let B = the RATE at which hose B can fill the pool alone

Let C = the RATE at which hose C can fill the pool alone

Hoses A and B working simultaneously can pump the pool full of water in 4 hours

From Question. The combined RATE of hoses A and B is 1/4 of the pool PER HOUR

In other words, A + B = ¼

Hoses B and C working simultaneously can pump the pool full of water in 6 hours

From Question. The combined RATE of hoses A and C is 1/4 of the pool PER HOUR

In other words, B + C = 1/6

Hoses A and C working simultaneously can pump the pool full of water in 5 hours

From Question. The combined RATE of hoses A and C is 1/5 of the pool PER HOUR

In other words, A + C = 1/5

At this point, we have the following system:

A + B = 1/4  ...1

B + C = 1/6  ...2

A + C = 1/5  ...3

Considering equation one

A  = 1/4 -B

A = (1 - 4B)/4 ....4

sub equation 4 into 2

(1 - 4B)/4 + C = 1/5

C = B - (1/20)

Considering equation  3

B + [B-(1/20)] = 1/6

B =7/80

Sub B Into enq 1 and 2

A =13/80

C = 3/80

A+B+C = 23/80

          3.47

=> 3 hours

0.47 × 60 = 28minutes

7 0
3 years ago
Read 2 more answers
Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a
Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }

Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}

Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

n = (16*(\frac{1.2}{(270-264)}))^2

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0
3 years ago
75% of 32 students are interested in Mathematics. How many are not interested in Mathematics ?
ValentinkaMS [17]

Answer:

The answer is 8!

Step-by-step explanation:

75% is 3/4 of 100 so you take 32÷4 and get 8, hope this helps:)

5 0
2 years ago
Read 2 more answers
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