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3 years ago

5. Draw the number line graph of -3.

Mathematics

1 answer:

lana [24]3 years ago

4 0

Answer:

9, 7,-8.3,>

Step-by-step explanation:

5)-3 is marked exactly 3 units to the left of origin 0.

(since negative sign we mark on left of 0)

6) |-9| = absolute value of 9 = 9

7) y-11=-4

Add 11 to both the sides

WE get y =7 is the solution

8) a + b when a = 2.5 and b = -10.8.

= 2.5+(-10.8) = -(10.8-2.5) = -8.3

9) 0.4 is positive hence lies to the right of origin.

Partition the space between 0 and 1 into 10 equal subparts and mark 4th subdivision to get 0.4

10) -1.8>-2.5

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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

Just question 7 thanks

ad-work [718]

Let x be the number of boys, the number of girls will be 200-x

Number of boys with blue eyes x/3
Number of girls with blue eyes (200-x)/4
ll blue eyes ==> x/3 + (200-x)/4 = 57
Reduce to same denominator & solve for x
(4x + 600 - 3x)/12 = 681/12
x=81, the total number of boys & girls200-81 119

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3 years ago

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