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Zigmanuir [339]
3 years ago
9

Find the value of x in each diagram.

Mathematics
2 answers:
sineoko [7]3 years ago
8 0

Answer:

6) 4 , 7) 35

Step-by-step explanation:

IgorLugansk [536]3 years ago
6 0

Answer:

x = 8 and x = 7

Step-by-step explanation:

(6)

Since the triangles are similar then the ratio of corresponding sides are equal, that is

\frac{ST}{MN} = \frac{RS}{LM} , substitute values

\frac{x}{4} = \frac{4}{2} ( cross- multiply )

2x = 16 ( divide both sides by 2 )

x = 8

(7)

Since the prisms are similar then the ratio of corresponding sides are equal, that is

\frac{x}{35} = \frac{2}{10} ( cross- multiply )

10x = 70 ( divide both sides by 10 )

x = 7

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What is the hight of a cone with a radius of 4 inches and of volume of 8 cubic inches? Round your Answer to the tenths place.
Scorpion4ik [409]

Answer:

0.48

Step-by-step explanation:

You use the formula V=πr2 h/3

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3 years ago
Rewrite using the distributive property then solve using gcf. 5(4+3)
Vitek1552 [10]

Answer:

Step-by-step explanation:

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For the standard normal probability distribution, the area to the right of the mean is:
sveticcg [70]
The answer is 0.5. I hope this is helpful.
7 0
3 years ago
The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont
BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)

and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so  A∩B = B and  

\bf P(A | B) = P(B)P(B) = (P(B))^2

We have  

\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237

hence,

\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection =  </em>

\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021

5 0
3 years ago
I don’t know if anyone will see you this but do not open a link that someone gives you for a answer
yKpoI14uk [10]

Answer:

ok thanks for ur care my friend have a wonderful day

7 0
3 years ago
Read 2 more answers
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