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4 years ago

PLS HELP ME GUYS I NEED HELP WITH GEOMETRY

Mathematics

1 answer:

Alekssandra [29.7K]4 years ago

6 0

Answer:

7, -21

Step-by-step explanation:

F could be located 14 to the left of -7 or 14 to the right of -7. 7 and -21 could be the possible locations of F.

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In a college library there are 4 times as many nonfiction books as fiction books. How many times the number of nonfiction books

balu736 [363]

Four? Isn't it stated in the question? Unless I am assuming wrong.

8 0

3 years ago

Read 2 more answers

A-t shirt cost $4 a sweter cost 4 times as much as the t-s

slava [35]

T-shirt costs $4
Sweater costs $4(4)

Sweater costs 4 x 4

4 x 4 = 16

Sweater costs $16

hope this helps

5 0

3 years ago

Read 2 more answers

A semi-circle window can be used above a doorway or as an accent window

anzhelika [568]

A semicircle is a part of a circle, and it is referred to as half of a given circle. Thud the area of the semi-circle window is 982 $in^{2}$ .

A circle is a shape that is bounded by a curved path which is referred to as the circumference. Some parts of a circle are radius, diameter, sector, arc, semi-circle, circumference, etc.

A semicircle is a part of a circle, and it is referred to as half of a given circle.

such that:

Area of a circle = $\pi$ $r^{2}$

and

area of a semicircle = $\frac{\pi r^{2} }{2}$

where: r is the radius of the circle, and $\pi$ is a constant with a value of $\frac{22}{7}$ .

Thus from the given question, it can be inferred that;

r = $\frac{50}{2}$

= 25

r = 25 in

Thus, the area of the semi-circle can be determined as;

area of the semi-circle = $\frac{1}{2}$ * $\frac{22}{7}$ * $25^{2}$

= 982.1429

area of the semi-circle = 982.14 $in^{2}$

The area of the semi-circle window is approximately 982 $in^{2}$ .

for more clarifications on the area of a semi-circle, visit: brainly.com/question/15937849

#SPJ 1

8 0

2 years ago

A number is squared, then doubled and then 4 is subtracted. The result is the original number. What could the number be?

s344n2d4d5 [400]

Answer:

umm its B cause i got it right hope this helped :)

Step-by-step explanation:

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