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3 years ago

Write a quadratic equation having the given solutions -8

Mathematics

1 answer:

Anna71 [15]3 years ago

6 0

Answer:

$y=x^2+16x+64$

Step-by-step explanation:

A quadratic needs 2 solutions, but since you only have 1, I am to assume that x = -8 twice (or multiplicity of 2). If x = -8, then in factor form, it's (x + 8). Muliplying that by itself twice:

(x + 8)(x + 8):

$x^2+8x+8x+64=0$

Simplifying gives you:

$x^2+16x+64=y$

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On December 1, Jasmin Ernst organized Ernst Consulting. On December 3, the owner contributed $85,360 in assets in exchange for i

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The income statement illustrates that the net income will be $3530.

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1 year ago

Place these real numbers in ascending order

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Answer:

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6 0

3 years ago

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

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

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

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

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

2 years ago

A soft-drink machine can be regulated so that it discharges an average of μ oz. per cup. If the ounces of fill are Normally dist

sattari [20]

Answer:

μ = 5.068 oz

Step-by-step explanation:

Normal distribution formula to use the table attached

Z = (x - μ)/σ

where μ is mean, σ is standard deviation, Z is on x-axis and x is a desired point.

98% of 6-oz. cups will not overflow means that the area below the curve is equal to 0.49; note that the curve is symmetrical respect zero, so, 98% of the cases relied between the interval (μ - some value) and (μ + some value)].

From table attached, area = 0.49 when Z = 2.33. From data, σ = 0.4 oz and x = 6 oz (maximum capacity of the cup). Isolating x from the formula gives

Z = (x - μ)/σ

2.33 = (6 - μ)/0.4

μ = 6 - 2.33*0.4

μ = 5.068

This means that with a mean of 5 oz and a standard deviation of 0.4 oz, the machine will discharge a maximum of 6 oz in the 98% of the cases.

4 0

3 years ago