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3 years ago

Find the unit rate.

Mathematics

1 answer:

xxMikexx [17]3 years ago

8 0

Answer:

22/40

Step-by-step explanation:

find the equivalent ratio of 4 and 10 and it is 40 so the then you multiply 4 times 10 and 10 times 4 then you multiply 1x10 and 10x4 so now your new problem will be

10/40+12/40 which is 22/40=1 8/15

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The heights of 10 year old children has a normal probability distribution with mean of 54.6 inches and standard deviation of 5.7

Fofino [41]

Answer:

a) 0.69

The probability that a randomly selected 10-year old child will be more than 51.75 inches tall

P(X>51.75 ) = 0.6915

Step-by-step explanation:

Step(i):-

Given mean of the Population = 54.6 inches

Given standard deviation of the Population = 5.7 inches

Let 'X' be the random variable of normal distribution

Let 'X' = 51.75 inches

$Z = \frac{x-mean}{S.D} = \frac{51.75-54.6}{5.7} = -0.5$

Step(ii):-

The probability that a randomly selected 10-year old child will be more than 51.75 inches tall

P(X>51.75 ) = P(Z>-0.5)

= 1 - P( Z < -0.5)

= 1 - (0.5 - A(-0.5))

= 1 -0.5 + A(-0.5)

= 0.5 + A(0.5) (∵A(-0.5)= A(0.5)

= 0.5 +0.1915

= 0.6915

Conclusion:-

The probability that a randomly selected 10-year old child will be more than 51.75 inches tall

P(X>51.75 ) = 0.6915

5 0

3 years ago

I dont get it....plzzz help me

dsp73

This is a no solution equation.

HOPE THIS HELPS!

6 0

3 years ago

Please solve it .........

VLD [36.1K]

Answer:

Step-by-step explanation:

We must first solve the first equation to solve the second question. If we make up an answer for each variable in the first equation, it will make it easier to solve the next. Now lets say that each of the fractions will be equal to 1/3. We can then find out that x=2, y=4, and z=4012. We then have to factor the second equation's fractions. we can factor out x, y, and z out of each fraction respectively. The equations become the same as the first set, but x, y, and z instead of 1, 2 and 2006. because we know that x=2, y=4, and z=4012, we can simplify each fraction to become 2/3, and add these fractions up to equal 2.

4 0

3 years ago

Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1,

OLEGan [10]

Let $D$ denote the event that an HD is defective, and $F_i$ the event that a particular HD was produced at facility $i$ .

You are asked to compute

$\mathbb P(F_2\mid D)$
$\mathbb P(F_4\mid D)$
$\mathbb P(D^C)$

From the definition of conditional probabilities, the first two will require that you first find $\mathbb P(D)$ . Once you have this, part (c) is trivial.

I'll demonstrate the computation for part (a). Part (b) is nearly identical.

(a)
$\mathbb P(F_2\mid D)=\dfrac{\mathbb P(F_2\cap D)}{\mathbb P(D)}$

Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so $\mathbb P(F_2\cap D)=\mathbb P(F_2)\mathbb P(D)=0.20\times0.015$ .

Use the law of total probability to determine the value of the denominator:

$\mathbb P(D)=\mathbb P(D\mid F_1)+\mathbb P(D\mid F_2)+\mathbb P(D\mid F_3)+\mathbb P(D\mid F_4)$

We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So

$\mathbb P(D)=0.015+0.02+0.01+0.03=0.075$

and thus

$\mathbb P(F_2\mid D)=\dfrac{0.2\times0.015}{0.075}=0.04$

(b) Similarly,
$\mathbb P(F_4\mid D)=\dfrac{0.4\times0.03}{0.075}=0.16$

(c)
$\mathbb P(D^C)=1-\mathbb P(D)=0.925$

6 0

3 years ago

It won't let me get passed unless I watch a video and the video doesn't work

allochka39001 [22]

Answer:

GO out and go back in maybe that will work

Step-by-step explanation:

3 0

4 years ago