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alex41 [277]
3 years ago
8

How many ways can a set of four tires be put on a car if all the tires are interchangeable? How many ways are possible if two of

the four are snow tires?
Mathematics
1 answer:
stellarik [79]3 years ago
4 0

Answer: 24 possible ways

A set of 4 tires can be fixed in any of the four possible positions in a car as it is mentioned that all four tires are interchangeable. Therefore, the 4 tires can be fixed in 4! ways. The four interchangeable tires can be put on a car in 24 possible ways.

Step-by-step explanation:

looked it up

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Write 0.0002839 in scientific notation.
Anika [276]

Answer:2.84 x 10-4

There you go!

8 0
4 years ago
Read 2 more answers
The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much
Y_Kistochka [10]

Answer:

x = ∛ 2*V/5  

y = ∛ 2*V/5

h  = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is  x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h      where   h =  V / x*y

As the base is a rectangular one there are 2 sides x*h .  and 2 sides  y*h

According to this:

Ct (cost of aquarium )  = cost of the base  + cost of the sides

cₐ  ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h   +   c₁*2*y*h

C(t)  =  5*c₁*x*y +2* c₁*x* V/x*y  +  2* c₁*y* V/x*y    or

C(t)  =  5*c₁*x*y  + 2*c₁*V/y   *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x)  =  5*c₁*y - 2*c₁*V/x²

C´(y)  =  5*c₁*x - 2*c₁*V/y²

C´(x)  = C´(y)        ⇒  5*c₁*y - 2*c₁*V/x²  =   5*c₁*x - 2*c₁*V/y²

or    5*y - 2*V/x²  =   5*x - 2*V/y²

(5*y*x² - 2*V)/x²  = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y²  = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y²  =  5*y²x³  - 2*V*x²

5*y³*x² - 5*y²x³  =  2*V * ( y² - x²)

by symmetry  x =  y

Then using x = y  and plugging that value on the derivatives

C´(x) =  5*c₁*y - 2*c₁*V/x²

C´(x) =  5*c₁*x - 2*c₁*V/x²

C´(x) = 0          ⇒     5*c₁*x - 2*c₁*V/x²  = 0

5*x  - 2*V/x² = 0      ⇒  5*x³ - 2*V = 0   ⇒   5*x³  = 2*V  ⇒ x³ = 2*V/5

x = ∛ 2*V/5       and   y = ∛ 2*V/5    and   h  =  V/ x*y    h  = V/ ∛ 4*V²/25

7 0
3 years ago
Evaluate the expression -434 + (-1.5)2
lorasvet [3.4K]

Answer: -437

Step-by-step explanation:

5 0
3 years ago
PLEASE HELP !!!!
My name is Ann [436]
1day=24 hours
10 days=240hours

60mins=1hr
30mins=1/2hr

total is 240hr+21hr+0.5hr=261.5hr


total is 261.5hr
7 0
3 years ago
To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil
Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

H_0: The true average penetration is 50 mils

H_a: The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the <em>level of confidence</em> is α = 0.05 then the z_\alpha score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)

The z-score associated with this test is

z=\frac{\bar x-\mu}{s/\sqrt{n}}

where

\bar x = <em>mean of the sample</em>

\mu = <em>average established by the specification</em>

s = <em>standard deviation of the sample</em>

n = <em>size of the sample</em>

Computing this value of z we get z = 3.42

Since z >z_\alpha we can conclude that the sample has not met the required specification.

5 0
4 years ago
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