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2 years ago

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Write an expression for the number of tiles Bruce used and an expression for the number of tiles Felicia used. Use x to represen

t the number of tiles in a box.

2 answers:

Basile [38]2 years ago

8 0

Expression for the number of tiles Bruce used : <u>5x+2</u>

Expression for the number of tiles Felicia used : <u>5x + 5</u>

<h3>Further Explanation </h3>

One variable linear equation is an equation that has a variable and the exponent number is one.

Can be stated in the form:

$\large{\boxed{\bold{ax=b}}$

or

ax + b = c, where a, b, and c are constants, x is a variable

We complete the introduction for the task

Bruce retiled his kitchen. Originally, he bought three boxes of tile with the same number of tiles in each. He ran out of tile and had to go back to get three extra boxes of tile. However, he only used two tiles from the last box to finish the job.

Felicia also retiled her kitchen. She bought five boxes of tile with the same number of tiles that were in each as Bruce’s boxes. She also ran out of tile. She had to go back to the store and get an extra five tiles to finish the job.

We can translate those in to expression of One variable linear equation (x = number of tiles)

Bruce

He bought three boxes ⇒ 3x

He only used two tiles from the last box ⇒ 2x + 2

Total tiles of Bruce used : 3x+2x+2 = 5x+2

Felicia

She bought five boxes ⇒ 5x

She get an extra five tiles to finish the job. ⇒ 5 tiles

Total tiles of Felicia used : 5x+5

<h3>Learn more</h3>

linear equation

brainly.com/question/13101704

Keywords : an expression, linear equation, tiles, box,Bruce,Felicia

#LearnWithBrainly

galina1969 [7]2 years ago

5 0

Answer:

bruce: 5x+2

Felicia: 5x+5

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Consider the trinomial $5x^2-2x-3.$

1. Find the discriminant:

$D=b^2-4ac=(-2)^2-4\cdot 5\cdot (-3)=4+60=64,\\ \\\sqrt{D}=8>0.$

2. Find the roots of given trinomial:

$x_{1,2}=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{2\pm 8}{2\cdot 5}=1,-0.6.$

3. Conclusion: given trinomial has two different real roots.

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Which are the solutions of x2 = –5x + 8? StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFract

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Answer:

$x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Step-by-step explanation:

Given:

The equation to solve is given as:

$x^2=-5x+8$

Rearrange the given equation in standard form $ax^2+bx +c =0$ , where, $a,\ b,\ and\ c$ are constants.

Therefore, we add $5x-8$ on both sides to get,

$x^2+5x-8=0$

Here, $a=1,b=5,c=-8$

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Plug in $a=1,b=5,c=-8$ and solve for $x$ .

$x=\frac{-5\pm \sqrt{5^2-4(1)(-8)}}{2(1)}\\x=\frac{-5\pm \sqrt{25+32}}{2}\\x=\frac{-5\pm \sqrt{57}}{2}\\\\\\\therefore x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Therefore, the solutions are:

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Answer:

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Step-by-step explanation:

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Mean = np

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So, Mean = n × p = (18) ×(0.30) = 5.4

Next we we will find the standard deviation:

Standard Deviation = $\sqrt{npq}$

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Now calculate the Z score for 6 successes.

Z = ( of successes we're interested in - Mean) ÷ (Standard Deviation)

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2 years ago

A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr

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Answer:

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No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

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The estimated standard error is:

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The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

$LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096$

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

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2 years ago

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

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Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

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We calculate the number of favorable combinations:

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Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

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