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2 years ago

PLEASE HELP ME I NEED THIS IN 15 MINUTES!!!

Mathematics

2 answers:

Vedmedyk [2.9K]2 years ago

6 0

The demensions are 4” wide 2” height and 12” length, forgive me if im wrong

algol [13]2 years ago

5 0

Answer:

4” wide 2” height and 12”

Step-by-step explanation:

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PLEASE HELP SUPER EASY 10 POINTS DUE TOMARROW

Volgvan

It would be about 16.67 for a regular price of the Vans?

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3 years ago

Distance between A (2, -1) and (2,a) is 4 units. Find the value of a where a is positive.

kolezko [41]

Answer:

a = 3

Step-by-step explanation:

Calculate the distance d using the distance formula and equate to 4

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (2, - 1) and (x₂, y₂ ) = (2, a)

d = $\sqrt{(2-2)^2+(a+1)^2}$ = $\sqrt{0+(a+1)^2}$ = $\sqrt{(a+1)^2}$ = a + 1 then

a + 1 = 4 ( subtract 1 from both sides )

a = 3

3 0

2 years ago

The table represents a liner function. What is the slope of the function?

Liono4ka [1.6K]

Answer:

The slope is 3.

Step-by-step explanation:

Let's use the slope formula to calculate the slope of this function. Remember, slope equals rise over run, or the difference between y coordinates divided by the difference between x coordinates of 2 points on the graph.

Let's use the last 2 points in the table: (1, 7) and (2, 10)

$\frac{y_{2} - y_{1} }{x_{2} - x_{1} }$

= $\frac{10 - 7}{2 - 1}$

= $\frac{3}{1}$

The slope of this function is 3!

Hope this helps :) Feel free to ask me any questions!

5 0

3 years ago

Read 2 more answers

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

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

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.

This means that,

Probability that at least one of them needs replacement at the first inspection =

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

5 0

3 years ago

Complete function tables part 2 please help i will mark as brainlyst

Vlad1618 [11]

1 = 1/2(0) + 1

2 = 1/2(2) + 1

3 = 1/2(4) + 1

4 = 1/2(6) + 1

hope this helps!

6 0

3 years ago