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3 years ago

(x + 2)(x + 3)(x + 4) can be written in the form x + ax² + bx + c

Mathematics

1 answer:

ELEN [110]3 years ago

5 0

Answer:

$a=9, b=26, c=24$

Step-by-step explanation:

Let's expand the expression bit by bit.

$(x+2)(x+3)=(x)(x)+(x)(3)+(2)(x)+(2)(3)=x^2+5x+6$ by FOIL.

We have that $(x^2+5x+6)(x+4)=x(x^2+5x+6)+4(x^2+5x+6)=x^3+5x^2+6x+4x^2+20x+24=x^3+9x^2+26x+24$ .

So, the answer is $\boxed{a=9, b=26, c=24}$

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The average amount of money spent for lunch per person in the college cafeteria is $6.35 and the standard deviation is $2.31. Su

alexdok [17]

Answer:

(6.35, 5.34)

(6.35, 0.99)

0.10253

0.3984

Step-by-step explanation:

Given that :

μ = 6.35

σ = 2.31

n = 41

What is the distribution of X?X ~ N(,)

X :

Mean of distribution = μ= 6.35

The variance = σ^2 = 2.31^2 = 5.3361 = 5.34

X ~ N = (6.35, 5.34)

What is the distribution of x¯? x¯ ~ N(,)

The mean = μ = 6.35

The standard deviation of the mean = σ/sqrt(n) = 6.35/sqrt(41) = 0.9917 = 0.99

X ~N = (6.35, 0.99)

find the probability that this patron's lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757)

Obtain the standardized scores:

Z = (x - μ) / σ ; (6.1605 - 6.35) / 2.31 = - 0.082

P(Z < - 0.082) = 0.46732 (Z probability calculator)

(6.757 - 6.35) / 2.31 = 0.176

P(Z < 0.176) = 0.56985 (Z probability calculator)

P(Z < 0.176) -P(Z < - 0.082)

0.56985 - 0.46732 = 0.10253

For the group of 17 patrons, find the probability that the average lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757) ; n = 17

Obtain the standardized scores:

Z = (x - μ) / σ/sqrt(n) ; (6.1605 - 6.35) / 2.31/sqrt(17) = - 0.338

P(Z < - 0.338) = 0.36768 (Z probability calculator)

(6.757 - 6.35) / 2.31/sqrt(17) = 0.726

P(Z < 0.726) = 0.76608 (Z probability calculator)

P(Z < 0.726) -P(Z < - 0.338)

0.76608 - 0.36768 = 0.3984

8 0

3 years ago

Jen is planning a trip to

ludmilkaskok [199]

Answer:

$1800-$350=$1450

$1450÷$145=10 nights

3 0

2 years ago

While Preparing For A Morning Conference, Principal Corsetti Is Laying Out 8 Dozen Bagels On Square Plates. Each Plate Can Hold

USPshnik [31]

We have 8 dozen bagels, or 8*12=96 bagels. Each plate can hold 14 bagels, so we have enough bagels to fill 96/14=about 6.86 plates. However, we cannot have a fraction of a plate, so we round up to have a total of seven plates. To fill all seven plates fully, 7*14=98 bagels would be needed, which is two more than we have.

To summarize, Mr. Corsetti has seven plates of bagels, and would need two more bagels to fill the last one up.

6 0

3 years ago

Meteorologists in Texas want to increase the amount of rain delivered by thunderheads by seeding the clouds. Without seeding, th

Dafna11 [192]

Answer:

There is no significant difference in the amount of rain produced when seeding the clouds.

Step-by-step explanation:

Assuming that the amount of rain delivered by thunderheads follows a distribution close to a normal one, we can formulate a hypothesis z-test:

Null Hypothesis

$\bf H_0$ : Average of the amount of rain delivered by thunderheads without seeding the clouds = 300 acrefeet.

Alternative Hypothesis

$\bf H_a$ : Average of the amount of rain delivered by thunderheads by seeding the clouds > 300 acrefeet.

This is a right-tailed test.

Our z-statistic is

$\bf z=\frac{370.4-300}{300.1/\sqrt{30}}=1.2845$

We now compare this value with the z-critical for a 0.05 significance level. This is a value $\bf z^*$ such that the area under the Normal curve to the left of $\bf z^*$ is less than or equal to 0.05

We can find this value with tables, calculators or spreadsheets.

In Excel or OpenOffice Calc use the function

NORMSINV(0.95)

an we obtain a value of

$\bf z^*$ = 1.645

Since 1.2845 is not greater than 1.645 we cannot reject the null, so the conclusion that can be drawn when the significance level is 0.05 is that there is no significant difference in the amount of rain produced when seeding the clouds.

8 0

3 years ago

PLEASE HURRRY!!!!!!!!!!!!!

kifflom [539]

22+2y

<h2>hope it helps you ❣❣ </h2>

Mark me as brainliest

4 0

3 years ago

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