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PolarNik [594]
2 years ago
10

NEED HELP I SUCK AT GEOMETRY (OFFERING 100POINTS)!!! best answer get brainlest

Mathematics
2 answers:
Andrej [43]2 years ago
8 0

Answer:

g = 3.5

f = 78°

<u>Finding</u><u> </u><u>g</u><u>:</u>

25 is parallel to 8g - 3, so they are equal to each other.

25 = 8g - 3

28 = 8g

g = 3.5

<u>Finding</u><u> </u><u>f</u><u>:</u>

72° create a sum of 180° when adding to (f + 30) because they are supplementary.

72 + f + 30 = 180°

102 + f = 180°

f = 78°

Hope this helps!

Akimi4 [234]2 years ago
6 0

Answer:

the required values are

f = 78°

g= 3.5

Step-by-step explanation:

here's the solution : -

in the given figure, its a parallelogram, so it's adjacent angles are supplymentary

So,

=》f + 30° + 72° = 180°

=》f + 102° = 180°

=》f = 180° - 102°

=》f = 78°

now, since it's a parallelogram it's opposite sides are equal so,

=》8g - 3 = 25

=》8g = 25 + 3

=》g = 28 ÷ 8

=》g = 3.5

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32 POINTS
Nimfa-mama [501]

Answer:

c

Step-by-step explanation:

5 0
3 years ago
Stealth Bank has deposits of $700 million. It holds reserves of $70 million and has purchased government bonds worth $215 millio
zhannawk [14.2K]

Answer:

$75 million

Step-by-step explanation:

Given that:

Reserve value = $70 million

Purchased government bond = 215 million

Market value of loan = $490 million

Net worth :

Assets - liability

Assets = (Market value of loan + purchased government bond + reserve value)

(490 million + 215 million + 70 million)

275 million + 70 million

= $775 million

Liability = Deposits = $700 million

Net worth = ($775 - $700) million

Net worth = $75 million

4 0
2 years ago
How many different ways can six people be arranged in a line?
lesantik [10]
There are six people which means that each person can be arranged in 6 various arrangements

6×6=36

Answer: There can be 36 different ways for 6 people to be arranged 


8 0
3 years ago
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
2 years ago
What is the square root of 36x to the fourth power?
-BARSIC- [3]
\sqrt{36x^4}=\sqrt{36}\cdot\sqrt{x^4}=\sqrt{6^2}\cdot\sqrt{(x^2)^2}=6\cdot x^2=\boxed{6x^2}
8 0
3 years ago
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