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3 years ago

Choosing brainliest if you can solve. show all work plz!! :))

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

5 0

Answer:

1. X=3 Y=2 Put at (3, 8)

I'm sorry I have to go but this is the first answer, good luck with the test! :)

Step-by-step explanation:

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Find the discriminant of 2x square + x - 6 is equal to zero?

Igoryamba

Answer:

When the discriminant value is zero, we get one real solution; When the discriminant value is negative, we get a pair of complex solutions; Standard Form. The standard discriminant form for the quadratic equation ax 2 + bx + c = 0 is. Discriminant, D = b 2 – 4ac. Where. a is the coefficient of x 2. b is the coefficient of x. c is a constant term.

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Step-by-step explanation:

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3 years ago

Minimum point of the quadratic x^2+6x-2

WITCHER [35]

Answer:

(-3,-11)

Step-by-step explanation:

Compare the given quadratic equation with the general quadratic equation.

a=1, b=6 and c=-2

$x=-\dfrac{(6)}{2(1)}\\=-3$

Subsitute $-3$ for $x$ in given quadratic equation.

$y=(-3)^2+6(-3)-2\\=9-18-2\\=-11$

The minimum point is (-3,-11).

3 0

3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$