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1 year ago

The perimeter of the rectangle below is:

Mathematics

1 answer:

blondinia [14]1 year ago

5 0

Answer:

18x + 28 is the perimeter

Step-by-step explanation:

As perimeter is the som of all side

So P = 6x+8 + 3x+6 + 6x+5 + 2x+2 + 3 + x+4

= 6x+3x+6x+2x+x + 8+6+5+2+3+4

= 18x + 28

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Graph the linear equation find three points that solves the equation 2y=5x+11

Julli [10]

Answer:

<h3>Attachment</h3>

Step-by-step explanation:

Convert to the slope-intercept form (<em>y = mx + b</em>):

$2y=5x+11$ <em>divide both sides by 2</em>

$\dfrac{2y}{2}=\dfrac{5x}{2}+\dfrac{11}{2}\\\\y=\dfrac{5}{2}x+\dfrac{11}{2}$

We choose any three x values and calculate the y value:

for x = -1

$y=\dfrac{5}{2}(-1)+\dfrac{11}{2}=-\dfrac{5}{2}+\dfrac{11}{2}=\dfrac{6}{2}=3\to A(-1,\ 3)$

for x = 1

$y=\dfrac{5}{2}(1)+\dfrac{11}{2}=\dfrac{5}{2}+\dfrac{11}{2}=\dfrac{16}{2}=8\to B(1,\ 8)$

for x = -3

$y=\dfrac{5}{2}(-3)+\dfrac{11}{2}=-\dfrac{15}{2}+\dfrac{11}{2}=-\dfrac{4}{2}=-2\to C(-3,\ -2)$

Mark the points on the coordinate plane and draw a line through the given points.

6 0

2 years ago

These box plots show the prices for two different brands of shoes.

nignag [31]

Answer: Choice C

The interquartile range (IQR) for brand A, $10, is less than the IQR for brand B, $20.

===========================================================

Explanation:

Let's go through the answer choices one by one to see which is true and which is false. Also, we'll see which is helpful when it comes to the spread.

----------------------

Choice A

The IQR is found by subtracting the values of Q1 and Q3, which are the left and right edges of the box in that order.

For boxplot A, we have an IQR = Q3-Q1 = 80-70 = 10.

So we can immediately rule out choice A because the IQR for boxplot A is not 20, but instead it's 10.

----------------------

Choice B

The statement made in choice B is true, since the center vertical lines represent the median, but it's not useful to determine the spread. The median only tells us the center of the distribution. It doesn't tell us how spread out the distribution is.

Imagine you didn't have access to the graph. Now if you're told that the median for brand A is larger than brand B's median, then you'd probably think boxplot A is more spread out; however, the boxplot diagram shows the opposite is the case.

So again the median isn't used here. We can rule out choice B.

----------------------

Choice C

This statement is true. We subtract the box edge values to get the IQR values for each boxplot

For brand A we have IQR = Q3-Q1 = 80-70 = 10

For brand B we have IQR = Q3-Q1 = 70-50 = 20

This is the answer we're after.

The IQR is one tool we can use to measure the spread. The range is another option. We could also use the variance or the standard deviation (two slightly similar ideas) to measure the spread.

As you can see from the diagram, the larger the IQR, the more spread out the data will be.

----------------------

Choice D

This is false because earlier we've shown 10 is the IQR for brand A.

7 0

2 years ago

You’re rectangular 12 feet long and 8 feet wide. Your friend’s rectangular vegetable garden 50 feet long and 10 feet wide are th

igor_vitrenko [27]

Answer:

Step-by-step explanation:

For shapes to be similar, their ratios of length-to-length and width-to-width must be the same. When you divide the larger length by the smaller length and the larger width by the smaller width, do you get the same number? No, so these rectangles are not similar.

5 0

2 years ago

Read 2 more answers

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0

2 years ago

MR MARK MARKS HIS CLASS ON A NORMAL CURVE. THOSE WITH z-SCORES ABOVE 1.8 WILL RECEIVE AN A, THOSE

lukranit [14]

Using the normal distribution, it is found that the percentages are given as follows:

3.59% of the grades will be A.

9.98% of the grades will be B.

74.92% of the grades will be C.

8.64% of the grades will be D.

2.87% of the grades will be F.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The prorportion of students who receive an A is <u>one subtracted by the p-value of Z = 1.8</u>.

Looking at the z-table, Z = 1.8 has a p-value of 0.9641.

1 - 0.9641 = 0.0359.

3.59% of the grades will be A.

For B, it is the <u>p-value of Z = 1.8 subtracted by the p-value of Z = 1.1</u>, hence:

0.9641 - 0.8643 = 0.0998.

9.98% of the grades will be B.

For C, it is the <u>p-value of Z = 1.1 subtracted by the p-value of Z = -1.2</u>, hence:

0.8643 - 0.1151 = 0.7492.

74.92% of the grades will be C.

For D, it is the <u>p-value of Z = -1.2 subtracted by the p-value of Z = -1.9</u>, hence:

0.1151 - 0.0287 = 0.0864.

8.64% of the grades will be D.

For F, it is the <u>p-value of Z = -1.9</u>, hence 2.87% of the grades will be F.

More can be learned about the normal distribution at brainly.com/question/15181104

#SPJ1

7 0

10 months ago