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adelina 88 [10]
4 years ago
7

A rectangle has side lengths of x + 3 and 2 – x. Find an expression that describes the area of the rectangle.

Mathematics
2 answers:
Rus_ich [418]4 years ago
5 0

Answer:

D


Step-by-step explanation:

We find the area of a rectangle by multiplying the lengths of both the sides <em>(length and width).</em>

  • One side given is x+3
  • Another side given is 2-x

Multiplying these 2 will give us the expression for the area of the rectangle <em>(we use distributive property given by  a(b+c)=ab+ac in this problem to multiply)</em>

Area = (x+3)(2-x)\\=2x-x^2+6-3x

<em>Now combining like terms and rearranging, we have:</em>

<em>2x-x^2+6-3x\\=-x-x^2+6\\=-x^2-x+6</em>


Answer choice D is right.


bezimeni [28]4 years ago
3 0

Answer:

D

Step-by-step explanation:

distributive property:

2x - x^2 + 6 - 3x

simplify:

-x^2 - x + 6


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The GCD(a, b) = 6, the LCM(a, b)=180. Find the least possible value of a+b.
julia-pushkina [17]
This is the concept of numbers, given that the G.C.D two numbers (a,b)=6 and the L.C.M (a,b)=180, to find the possible number we do as follows;
L.C.M =a*b=180
since 6 is a multiple of a and b, then the numbers could be:
180/6
=30
hence these numbers could be (6,30)
5 0
3 years ago
A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 47.0 and 52.
Pavlova-9 [17]

Answer:

0.15 or 15%

Step-by-step explanation:

Since lengths are uniformly distributed, the probability that a class period runs between a two exact times is:

P(x_1\leq x \leq x_2)=\frac{x_2-x_1}{b-a}

In this case, a = 47.0 and b = 52.0 minutes.

The probability that a given class period runs between 50.25 and 51.0 minutes is:

P(50.25\leq x \leq 51.0)=\frac{51.0-50.25}{52-47} \\P(50.25\leq x \leq 51.0)=0.15=15\%

The probability is 0.15 or 15%.

7 0
3 years ago
Im bad at math:( can somone please help
Mazyrski [523]
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8 0
3 years ago
What is the formula for a confidence interval?
ikadub [295]

Answer:

a) The formula is given by mean \pm the margin of error. Where the margin of error is the product between the critical value from the normal standard distribution at the confidence level selected and the standard deviation for the sample mean.

b) \bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

If the distribution for X is normal or if the sample size is large enough we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Part a

The formula is given by mean \pm the margin of error. Where the margin of error is the product between the critical value from the normal standard distribution at the confidence level selected and the standard deviation for the sample mean.

Part b

The confidence interval for the mean is given by the following formula:  

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

5 0
4 years ago
15 players for a softball team show up for a game: (a) How many ways are there to choose 10 players to take the field? 3003 (b)
prohojiy [21]

Answer:

(a) 3003 ways

(b) 10897286400 ways

(c) 3002 ways

Step-by-step explanation:

Given

n = 15 --- 15 players

Solving (a) Ways of selecting 10 players.

This implies combination.

So, we have:

r=10

Using:

^nC_r = \frac{n!}{(n-r)!r!}

We have:

^{15}C_{10} = \frac{15!}{(15-10)!10!}

^{15}C_{10} = \frac{15!}{5!10!}

Simplify

^{15}C_{10} = \frac{15*14*13*12*11*10!}{5!10!}

^{15}C_{10} = \frac{15*14*13*12*11}{5!}

^{15}C_{10} = \frac{15*14*13*12*11}{5*4*3*2*1}

^{15}C_{10} = \frac{360360}{120}

^{15}C_{10} = 3003

Solving (b) Ways of assigning positions to 10 players.

This implies permutation.

So, we have:

r=10

Using:

^nP_r = \frac{n!}{(n-r)!}

We have:

^{15}P_{10} = \frac{15!}{(15-10)!}

^{15}P_{10} = \frac{15!}{5!}

Solve each factorial

^{15}P_{10} = \frac{1307674368000}{120}

^{15}P_{10} = 10897286400

Solving (c) Ways of choosing at least 1 woman

We have:

Men = 10

Women = 5

Ways of selecting 10 players is: (a) 3003 ways

Since the number of men are 10, there is 1 way of selecting 10 men (i.e. selection without women)

Using complement rule:

At least 1 woman = Total - No woman

At\ least\ 1\ woman = 3003 - 1

At\ least\ 1\ woman = 3002

5 0
3 years ago
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