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g100num [7]
2 years ago
7

In this Figure , find the area of the shaded region.​

Mathematics
1 answer:
Sonja [21]2 years ago
3 0
The answer is .5 cm. I hope it’s right
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Express the interval using inequality notation.<br> (-♾,-11) U (7,♾)
gulaghasi [49]

The chosen topic is not meant for use with this type of problem. Try the examples below.

[-1,9)u(2,10]

(−1,2)∪(−4,0)

(−1,29)∪(26,50)

4 0
3 years ago
Can someone help me on this one Please
Mumz [18]
X-4y=1
2x+y=11
ok so you have x-4y=1 and 2x+y=11
3x-3y=12 is your answer
6 0
3 years ago
Let a, b, c, and d be non-zero real numbers. If the quadratic equation ax(cx + d) = -b(cx+d) is solved for x, which of the follo
LiRa [457]

Answer:

Option d

Step-by-step explanation:

given that a, b, c, and d be non-zero real numbers.

ax(cx + d) = -b(cx+d) \\acx^2+x(ad+bc)+bd =0

we can factorise this equation by grouping

(acx^2+xad)+)xbc+bd) =0\\ax(cx+d) +b(cx+d) =0\\(ax+b)(cx+d) =0

Equate each factor to 0 to get

x=\frac{-b}{a} , \frac{-d}{c}

Ratio of one solution to another would be

\frac{-b}{a} / \frac{-d}{c} \\=\frac{ad}{bc}

So ratio would be ad/bc

Out of the four options given, option d is equal to this

So option d is right

4 0
3 years ago
Name four things that you can find on both a dot plot and box plot.
o-na [289]

Answer:Dots, Lines, Box's, and Circles.

Step-by-step explanation: You're welcome

8 0
3 years ago
The accompanying data represent the miles per gallon of a random sample of cars with a​ three-cylinder, 1.0 liter engine.a. Comp
Kobotan [32]

Answer:

1). Z score = 1.464733

Mean = 38.87917

Standard deviation = 3.609405

2). Quartiles:

Q1 = 37.025

Q2 = 38.450

Q3 = 40.800

3). IQR = 3.775

4).

CI (lower fence) = 37.4351

CI (upper fence) =  40.32323

5). There is outlier in the data set. Please see attached box plot for evidence.

Step-by-step explanation:

1). By Z score, we mean:

Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}},

where:

x-bar ==> mean(g) = 38.87917

\mu = 37.80

standard deviation = 3.609405

sample size (n) = 24.

2). By quantile, we mean:

Q = L + (i*(n/4) - Cf)*c

Where L is the lower class limit of the quartile class

Cf is the cumulative frequency before the quartile class

c  is the class size.

3) . IQR = Q3 - Q1

4). CI = \mu \pm *Z_{\alpha/2} \frac{\sigma}{\sqrt{n}},

Where Z_{\alpha/2}  ==> 1.96

In order to replicate and obtain the result, please use the R code below:

g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,  

41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)

boxplot(g)

Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))

mean(g)

quantile(g)

IQR(g)

CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))

CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))

4 0
3 years ago
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