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Ilia_Sergeevich [38]
3 years ago
6

What is the greatest common factor between 120 and 88?

Mathematics
2 answers:
Mice21 [21]3 years ago
7 0
The gcf of 88 and 120 is 8.
Anestetic [448]3 years ago
7 0

120:    2 2 2 3 5   88:    2 2 2     11 GCF:    2 2 2       The Greates Common Factor (GCF) is:   2 x 2 x 2 = 8

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Janey bought a used bicycle for $640. the bike was 3 years old when she bought it, and cost $850 new. assuming it decreases in v
Radda [10]
The first thing we must do in this case is to define the variables.
 We have then:
 v: the value of the bicycle
 t: years
 The standard equation of the line is:
 v-vo = m (t-to)
 The slope is:
 m = (v2-v1) / (t2-t1)
 Substituting values:
 m = (640-850) / (3-0)
 m = -70
 We choose an ordered pair:
 (to, vo) = (0, 850)
 Substituting values:
 v-850 = -70 (t-0)
 Rewriting:
 v = -70t + 850
 Answer:
 
assuming it decreases in value by the same amount each year, a linear equation for the value of the bicycle, v, when it is t years old is:
 
v = -70t + 850
6 0
4 years ago
there are 27 children in Ms. kloot's class. when the flu was going around , about 26% of the students in the class were absent.
ValentinkaMS [17]

Answer:

About 8 children were sick

Step-by-step explanation:

The total number of children in Ms. kloot's class is 27.

The percentage of students who were absent due to the flu 26.

Number of children who were sick

=  \frac{26}{100}  \times 27

This simplifies to:

= 7.02

Therefore about 8 children were sick

8 0
3 years ago
Can some one can do both​
patriot [66]
A) m=-3
B) n=-17
hope this helps x
would love to have brainliest if it is right :)
8 0
3 years ago
Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha
const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

P(1.83\leq X\leq  a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06

Thus, the value of <em>a</em> is 12.06.

7 0
3 years ago
Find the GCF of the list of terms<br> a(b+c), 5(b+c)
Paladinen [302]

a(b+c) and 5(b+c)


Answer: GCF = (b+c)


8 0
4 years ago
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