Ask question

Ask question

All categories

3 years ago

7

If the measure of FGH IS 64º and the measure of EGD is (4x+8), what is the value of x?

1 answer:

Ugo [173]3 years ago

7 0

$\\ \sf\longmapsto 4x+8=64$

$\\ \sf\longmapsto 4x=64-8$

$\\ \sf\longmapsto 4x=56$

$\\ \sf\longmapsto x=\dfrac{56}{4}$

$\\ \sf\longmapsto x=14$

You might be interested in

How to change 611/17 to improper fractions

Annette [7]

This is the answer ....

3 0

4 years ago

Read 2 more answers

help me again please, i have 4 math quizes to do and I'm horrible at math and right now my grade in math is 2 approching so I ne

topjm [15]

Answer: What was the question?

Step-by-step explanation: Was it when did I drink the most juice?

8 0

3 years ago

Read 2 more answers

Roger made 8 1/2 quarts of hot chocolate. Each mug holds 1/2 of a quart. How many mugs will Roger be able to fill?

blondinia [14]

Number of mugs = 8.5/0.5=17

7 0

3 years ago

Mylia has 20.65 m of wire. She used 15.8 m to make a sculpture. She cut the rest of the wire into 3 equal parts. What is the len

deff fn [24]

Answer:

1.6 meters

Step-by-step explanation:

20.65-15.8=4.85

4.85/3=1.6

5 0

3 years ago

The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

4 0

3 years ago