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4 years ago

Help! ASAP!!!

Mathematics

1 answer:

solniwko [45]4 years ago

6 0

Answer:

7.2 units

Step-by-step explanation:

Coordinates of city A = (0, 0)

Coordinates of city B = (6, 4)

Use the distance formula to find the distance between the two cities as follows:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$A(0, 0) = (x_1, y_1)$

$B(6, 4) = (x_2, y_2)$

Plug in the values into the formula

$AB = \sqrt{(6 - 0)^2 + (4 - 0)^2}$

$AB = \sqrt{(6)^2 + (4)^2}$

$AB = \sqrt{36 + 16} = \sqrt{52}$

$AB = 7.2$ (nearest tenth)

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Use quadratic formula solve .7x^2 - 28x - 7

beks73 [17]

Answer:

x = 2 + sqrt(5) or x = 2 - sqrt(5)

Step-by-step explanation using the quadratic formula:

Solve for x over the real numbers:

7 (x^2 - 4 x - 1) = 0

Divide both sides by 7:

x^2 - 4 x - 1 = 0

Add 1 to both sides:

x^2 - 4 x = 1

Add 4 to both sides:

x^2 - 4 x + 4 = 5

Write the left hand side as a square:

(x - 2)^2 = 5

Take the square root of both sides:

x - 2 = sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

x = 2 + sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

Answer: x = 2 + sqrt(5) or x = 2 - sqrt(5)

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4 years ago

erik drove 439.92 miles on 15.6 gallons of gas. To the nearest hundredth, how many miles could he drive on 39.7 gallons of gas?

Thepotemich [5.8K]

In the question it is already given that Eric drove for 439.92 miles on 15.6 gallons. It is required to find the distance traveled on 39.7 gallons of gas. Also we have to find the answer to the nearest hundredth.
Then,
In 15.6 gallons Eric can drive for a distance = 439.92 miles
In 39.7 gallons Eric can drive for a distance = [(439.92/15.6) * 39.7] miles
= 1119.54 miles
So the total distance traveled by Eric is 1120 miles with 39.7 gallons of gas. The answer has been calculated to the nearest hundredth.

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4 years ago

According to government data, 20% of employed women have never been married. If 10 employed women are selected at random, what i

Ierofanga [76]

Answer:

a) $P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

b) $P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

c) For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

And replacing we got:

$P(X\geq 8)=0.0000779$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Bin (n=10 ,p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

Let X the random variable "number of women that have never been married" , on this case we now that the distribution of the random variable is:

$X \sim Binom(n=10, p=0.2)$

Part a

We want to find this probability:

$P(X=2)$

And using the probability mass function we got:

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

Part b

For this case we want this probability:

$P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

We can find the individual probabilities and we got:

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

Part c

For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

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Answer:

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Step-by-step explanation:

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