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3 years ago

PLEASE HELP, WILL MARK BRAINLIEST

Mathematics

1 answer:

KatRina [158]3 years ago

4 0

Answer:

(y-0)=-3(x-2)

Step-by-step explanation:

im not sure if this os correct but i tried

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Plz help ASAP!!!!!!!!!! will mark brainliest

alekssr [168]

Answer:

X = 0.9

Step-by-step explanation:

$\frac{2.7}{5.1} = \frac{x}{1.7} \\ \frac{2.7 \times 1.7}{5.1} = x \\ x = 0.9$

5 0

3 years ago

Read 2 more answers

Need answer quickly. Thanks

ratelena [41]

Answer:

sopa de macaco

Step-by-step explanation:

macacoooooooo

8 0

3 years ago

Please help I will give 100 points

horsena [70]

Answer:

FGE = 180 - (14 + 78)°

= 88°

so angle GEC = 88°

4x + 88 = 180

4x = 180 - 88

4x = 92

x = 92/4

= 23

Step-by-step explanation:

4 0

2 years ago

For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger

11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let X = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = $p=\frac{1}{120}$ .

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...$

In this case we need to compute the probability of 1 or more than 1 catastrophes in n missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}$

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166$

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410$

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803$

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419$

6 0

3 years ago

Raju bought an old car for Rs.125000 and spent Rs.25000 on its repair . He sold the car for Rs.200000 . Find his gain or loss pe

Keith_Richards [23]

Answer: $33.33\%$

Step-by-step explanation:

Given

Raju buy an old car for $Rs\ 1,25,000$

He spent $Rs\ 25000$ on its repair

The selling price of the car $Rs\ 2,00,000$

So, the cost price is

$\Rightarrow C.P.=1,25,000+25,000=Rs\ 1,50,000$

Here, $S.P.> C.P.\quad \text{i.e. gain}$

Gain percent is

$\Rightarrow \text{Gain percent}=\dfrac{2,00,000-1,50,000}{1,50,000}\times 100=\dfrac{50,000}{1,50,000}\times 100\\\Rightarrow \text{Gain percent}=33.33\ \%$

6 0

3 years ago