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11111nata11111 [884]

3 years ago

10

Joseph has started completing the square on the equation 3x2 - 7x + 12 = 0. He has worked to the point where he has the expressi

on x2 - x = -4. Use complete sentences describe Joseph’s steps up to this point and whether or not his work is accurate.

1 answer:

MrMuchimi3 years ago

7 0

Joseph has started completing the square on the equation where

given equation is

$3x^2-7x+12=0$

move constant term to other side.

$3x^2-7x=-12$

divide both sides by coefficient of x^2 that is by 3

$x^2-\frac{7}{3}x=-\frac{12}{3}$

$x^2-\frac{7}{3}x=-4$

This is the same step upto which Joseph has worked. But he got different equation

$x^2-x=-4$

Hence his work is not accurate.

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Leakage from underground gasoline tanks at service stations can damage the environment. It is estimated that 25% of these tanks

Gnesinka [82]

Answer:

a) 3.75

b) 23.61% probability that fewer than 3 tanks will be found to be leaking

c) 0% the probability that at least 600 of these tanks are leaking

Step-by-step explanation:

For each tank there are only two possible outcomes. EIther they leak, or they do not. The probability of a tank leaking is independent of other tanks. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

To solve question c), i am going to approximate the binomial distribution to the normal.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

It is estimated that 25% of these tanks leak.

This means that $p = 0.25$

15 tanks chosen at random

This means that $n = 15$

a.What is the expected number of leaking tanks in such samples of 15?

$E(X) = np = 15*0.25 = 3.75$

b.What is the probability that fewer than 3 tanks will be found to be leaking?

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361$

23.61% probability that fewer than 3 tanks will be found to be leaking

c.Now you do a larger study, examining a random sample of 2000 tanks nationally. What is the probability that at least 600 of these tanks are leaking?

Now we have n = 2000. So

$\mu = E(X) = np = 2000*0.25 = 500$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.25*0.75} = 19.36$

This probability is 1 subtracted by the pvalue of Z when X = 600. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{600 - 500}{19.36}$

$Z = 5.16$

$Z = 5.16$ has a pvalue of 0.

0% the probability that at least 600 of these tanks are leaking

4 0

3 years ago

Solve for x. -0.65 + 0.45x = 5.4

FinnZ [79.3K]

Answer:

x = 13.444 ... (repeating decimal) or;

x = 13 4/9

Step-by-step explanation:

-0.65 + 0.45x = 5.4

First we will add 0.65 to each 'side' of the equation.

0.65 + (-0.65) + 0.45x = 0.65 + 5.4

0.65 + (-0.65) = 0

0.65 + 5.4 = 6.05

Input the numbers, and now the equation looks like this:

0 + 0.45x = 6.05

Remove the zero:

0.45x = 6.05

Divide each 'side' by 0.45.

0.45x ÷ 0.45 = 6.05 ÷ 0.45

0.45x ÷ 0.45 = 1x = x

6.05 ÷ 0.45 = 13.444 ... repeating decimal

Input the numbers, and now the equation looks like this:

x = 13.444 ... repeating decimal

We can also write this as:

x = 13 4/9 in fraction form

3 0

3 years ago

The rate (gallons per day) at which a pond loses water due to evaporation from one day after observation has started is give by

Andreyy89

Answer: see table below

<u>Step-by-step explanation:</u>

$\left\begin{array}{c|l}T&\qquad w(t)\\1&\dfrac{1}{1^2}=1\implies 1.00\\\\2&\dfrac{1}{2^2}=\dfrac{1}{4}\implies 0.25\\\\10&\dfrac{1}{10^2}=\dfrac{1}{100}\implies 0.01\\\\50&\dfrac{1}{50^2}=\dfrac{1}{2500}\implies 0.0004\\\\100&\dfrac{1}{100^2}=\dfrac{1}{10000}\implies 0.0001\\\\200&\dfrac{1}{200^2}=\dfrac{1}{40000}\implies 0.000025\\\\500&\dfrac{1}{500^2}=\dfrac{1}{250000}\implies 0.000004\\\\1000&\dfrac{1}{1000^2}=\dfrac{1}{100000}\implies 0.000001\\\end{array}\right$

4 0

3 years ago

Dwight deposits $150 into his new savings account. The account earns 5% interest compounded annually.

GrogVix [38]

Answer:

$A=\$150(1.05)^{t}$

Step-by-step explanation:

we know that

The compound interest formula for this problem is equal to

$A=P(1+r)^{t}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods in years

in this problem we have

$P=\$150\\ r=5\%=0.05$

substitute in the formula above

$A=\$150(1+0.05)^{t}$

$A=\$150(1.05)^{t}$

8 0

3 years ago

PLEASE HELP!! I WILL AWARD BRAINLIEST AND 50 POINTS!!!!!!!!!!!!!!!!!!!!!! For the given picture: a) List all of the triangles th

PIT_PIT [208]

Answer:

There are 4 triangles. For the exterior and remote interior, I’m naming the angles with three letters. If your teacher just wants one, it would be the middle one. (I assume they’ll want all three, since the letters just name points, not angles.)

1) A) Name of triangle: ABO

B) Exterior AngleS: EAO, BOC and AOD

C) Remote Interior Angles: For EAO, it’s ABO and AOB. For BOC, it’s BAO and ABO. For AOD, it’s BAO and ABO

2) A) Name of triangle: ABC

B) Exterior Angles:EAC

C) Remote Interior Angles: ABO and BCO

3) A) Name of triangle: DOC

B) Exterior Angles: ODH, DOA, COB, and OCG

C) Remote Interior Angles: For ODH, it’s DOC and COD. For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO. For OCG, it has CDO and DOC

4) A) Name Of triangle: COB

B) Exterior Angles: AOB, DOC, and OBF

C) Remote Interior Angles: For AOB, it’s OBC and OCB. For DOC, it’s OBC and OCB. For OBF, it’s BOC and OCB

Step-by-step explanation:

There are 4 triangles. For the exterior and remote interior, I’m naming the angles with three letters. If your teacher just wants one, it would be the middle one. (I assume they’ll want all three, since the letters just name points, not angles.)

1) A) Name of triangle: ABO

B) Exterior AngleS: EAO, BOC and AOD

C) Remote Interior Angles: For EAO, it’s ABO and AOB. For BOC, it’s BAO and ABO. For AOD, it’s BAO and ABO

2) A) Name of triangle: ABC

B) Exterior Angles:EAC

C) Remote Interior Angles: ABO and BCO

3) A) Name of triangle: DOC

B) Exterior Angles: ODH, DOA, COB, and OCG

C) Remote Interior Angles: For ODH, it’s DOC and COD. For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO. For OCG, it has CDO and DOC

4) A) Name Of triangle: COB

B) Exterior Angles: AOB, DOC, and OBF

C) Remote Interior Angles: For AOB, it’s OBC and OCB. For DOC, it’s OBC and OCB. For OBF, it’s BOC and OCB

6 0

3 years ago