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gogolik [260]
3 years ago
5

Estimate the difference between 206 and 167

Mathematics
2 answers:
svp [43]3 years ago
7 0
But if you estimate bit it could be 200-160

= 40
Dimas [21]3 years ago
3 0
Its a 39 difference 206-167= 39
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is picking out some movies to rent, and he has narrowed down his selections to 5 documentaries, 7 comedies, 4 mysteries, and 5 h
pogonyaev

Answer: 91

Step-by-step explanation:

Given : The number of documentaries = 5

The number of comedies = 7

The number of mysteries = 4

The number of horror films =5

The total number of movies other than comedy = 14

Now, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is given by :-

^7C_7\times^{14}C_2\\\\\dfrac{7!}{7!(7-7)!}\times\dfrac{14!}{2!(14-2)!}\\\\=(1)\times\dfrac{14\times13}{2}\\\\=91

Therefore, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is 91 .

5 0
3 years ago
A metal conduit will be used as a pathway for wiring through a concrete block. The conduit is a 5 foot long rod with an outer di
kolbaska11 [484]

Answer:  586.59 cubic centimeters .

Step-by-step explanation:

As per given . we have

Inner diameter =  1.8 inches

⇒Inner radius :r = 0.9 in.  (radius is half of diameter)

= 0.9 x (2.54) = 2.286 cm  [∵ 1 in . = 2.54 cm]

Outer diameter = 2 inches

⇒Outer radius : R = 1 inch  = 2.54 cm

Height : h = 5 feet = 5 x(30.48) = 152.4 cm   [∵  1 foot = 30.48 cm]

The formula to find the volume of a hollow cylinder :

V=\pi(R^2-r^2)h , where R= outer radius , r= inner radius and h= height.

Now , the volume of metal in the conduit :

V=(3.14)(( 2.54)^2-(2.286)^2)(152.4)

V=(3.14)(6.4516-5.225796)(152.4)

V=(3.14)(1.225804)(152.4)

V=586.591342944\approx586.59\ cm^3

Hence, the volume of metal in the conduit is 586.59 cubic centimeters .

5 0
3 years ago
A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a
Diano4ka-milaya [45]

Answer:  (C) 0.1591

Step-by-step explanation:

Given : A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with

\mu=42,100\text{ miles}

\sigma=2,510\text{ miles}

Let x be the random variable that represents the lifetime of the tires .

z-score : z=\dfrac{x-\mu}{\sigma}

For x= 44,500 miles

z=\dfrac{44500-42100}{2510}\approx0.96

For x= 48,000 miles

z=\dfrac{48000-42100}{2510}\approx2.35

Using the standard normal distribution table , we have

The p-value : P(44500

P(z

Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles =  0.1591

3 0
3 years ago
2x + 5y = 10<br>5y<br>x +<br>2.<br>+<br>= 4​
elena-s [515]
Could you attach photo of a problem. It appears 2+..... =4 is missing
4 0
3 years ago
Can someone help me with this question?
KonstantinChe [14]
-3. Put the two equation equal to each other and solve. 
4x+18=2x+12
-2x=-2x 
2x+18=12 
-18=-18 
2x=-6 
2x/2=-6/2
x=-3
8 0
3 years ago
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