Help me out plz geometry

I need to find the resultant matrix

fomenos

the answers are there

i hope it will help you

7 0

3 years ago

Someone pls help me I turn this in today the answer is there I just need to show work pls help me

dolphi86 [110]

Hello and Good Morning/Afternoon

Let's tale this problem step-by-step:

What does the problem want:

$\hookrightarrow \text{the unemployment compensation}$

Employment compensation =55% of the average of the last 26-week salary

$\hookrightarrow \text{Average of Natalia's salary} = (\text{all the salaries added up}) / (\text{number of weeks})\\\hookrightarrow \text{Average of Natalia's salary} = 18940 / 26 = 728.46 \text{ (Rounded down)}$

⇒ 55% of that average

$\hookrightarrow \frac{55}{100} *728.46 = 400.65$

Answer: $400.65

Hope that helps!

#LearnwithBrainly

5 0

1 year 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

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

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

(8y²)(-3x²y²)(2/3xy⁴) HELP PLEASEEEEEEEE

IRINA_888 [86]

Step-by-step explanation:

1 Remove parentheses.

8{y}^{2}\times -3{x}^{2}{y}^{2}\times \frac{2}{3}x{y}^{4}

8y

2

×−3x

2

y

2

×

3

2

xy

4

2 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}

b

a

×

d

c

=

bd

ac

.

\frac{8{y}^{2}\times -3{x}^{2}{y}^{2}\times 2x{y}^{4}}{3}

3

8y

2

×−3x

2

y

2

×2xy

4

3 Take out the constants.

\frac{(8\times -3\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

(8×−3×2)y

2

y

2

y

4

x

2

x

4 Simplify 8\times -38×−3 to -24−24.

\frac{(-24\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

(−24×2)y

2

y

2

y

4

x

2

x

5 Simplify -24\times 2−24×2 to -48−48.

\frac{-48{y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

−48y

2

y

2

y

4

x

2

x

6 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

a

x

b

=x

a+b

.

\frac{-48{y}^{2+2+4}{x}^{2+1}}{3}

3

−48y

2+2+4

x

2+1

7 Simplify 2+22+2 to 44.

\frac{-48{y}^{4+4}{x}^{2+1}}{3}

3

−48y

4+4

x

2+1

8 Simplify 4+44+4 to 88.

\frac{-48{y}^{8}{x}^{2+1}}{3}

3

−48y

8

x

2+1

9 Simplify 2+12+1 to 33.

\frac{-48{y}^{8}{x}^{3}}{3}

3

−48y

8

x

3

10 Move the negative sign to the left.

-\frac{48{y}^{8}{x}^{3}}{3}

−

3

48y

8

x

3

11 Simplify \frac{48{y}^{8}{x}^{3}}{3}

3

48y

8

x

3

to 16{y}^{8}{x}^{3}16y

8

x

3

.

-16{y}^{8}{x}^{3}

−16y

8

x

3

Done

6 0

2 years ago

(x+5)(y+10) what is the answer to this equation?

netineya [11]

(x + 5)(y + 10)
x(y + 10) + 5(y + 10)
x(y) + x(10) + 5(y) + 5(10)
xy + 10x + 5y + 50

7 0

2 years ago