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3 years ago

6x + 4y > 12 What’s the x and y intercepts

Mathematics

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

x interprets: 2,0

y interprets: 0,3

Step-by-step explanation:

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A pair of jeans cost $32 last year this year they cost $38 what was the percent increase? PLEASE ANSWER NEED HELP!

garik1379 [7]

Answer:

18.75%

Step-by-step explanation:

percent increase = $\frac{increase}{original cost}$ × 100%

Increase in cost = $38 - $32 = $6

percent increase = $\frac{6}{32}$ × 100% = $\frac{6(100)}{32}$ = 18.75%

5 0

3 years ago

These are all related to geometry and I need help!

nydimaria [60]

SOH-CAH-TOA

sin=opp/hyp

cos=adj/hyp

tan=opp/adj

1. Sin c=opp/hyp 2. tan=opp/adj 3. sin=opp/hyp

sin c=8/17 tan38°=x/16 sin38=18/x

x=18/sin38; x=29.236846

4 0

3 years ago

Omar grouped the terms and factored the GCF out of the groups of the polynomial 3x3 – 15x2 – 4x + 20. His work is shown.

Phoenix [80]

Answer:

2nd option

Step-by-step explanation:

Given

3x³ - 15x² - 4x + 20

step 1 ( group the first/second and third/fourth terms )

(3x³ - 15x² ) + (- 4x + 20)

step 2 ( factor each group )

3x² (x - 5) - 4(x - 5) ← note factor of - 4 ( not + 4 )

step 3 ( factor out (x - 5) from each term )

(x - 5)(3x² - 4)

4 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 =

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