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2 years ago

10

what is the length of a picture frame whose width is 3 inches and whose proportions are the same as as a 9-inch wide by 15-inch

long picture frame

1 answer:

forsale [732]2 years ago

3 0

Answer:

5"

Step-by-step explanation:

Set this up as a proportion: 3 is your constant of proportionality

3 / 5

9 / 15

3/5 x 3/3 = 9/15

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Without properly working spark plugs, a vehicle will not run. For a specific vehicle, the spark plugs are supposed to have a gap

Salsk061 [2.6K]

Answer:

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3) = 0.9922</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the mean of the Normal distribution = 4.1mm</em>

<em>Given that the standard deviation of the Normal distribution = 0.0075mm</em>

<em>Let 'X' be the random variable in a normal distribution</em>

Given that X₁ = 3.9mm

$Z_{1} = \frac{X_{1} -mean}{S.D}$

$Z_{1} = \frac{3.9 -4.1}{0.075} = -2.666$

Given that X₂ = 4.3mm

$Z_{2} = \frac{X_{2} -mean}{S.D}$

$Z_{1} = \frac{4.3 -4.1}{0.075} = 2.666$

<u><em>Step(ii):-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3) = P(-2.666≤Z≤2.666)</em>

<em> = P(Z≤2.666)-P(Z≤-2.666)</em>

<em> = 0.5 +A(2.666) - (0.5-A(2.666)</em>

<em> = 2 × A(2.666)</em>

<em> = 2×0.4961</em>

<em> = 0.9922</em>

<u><em>Final answer:-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3) = 0.9922</em>

<em></em>

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1 year ago

A helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. Angle ABC

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Please consider the attached graph.

We have been given that a helicopter flies 8 km due north from A to B. It then flies 5 km from B to C and returns to A as shown in the figure. The measure of angle ABC is 150°. We are asked to find the area of triangle ABC.

We will use trigonometric area formula to solve our given problem.

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Answer:

A and C

Step-by-step explanation:

To determine which events are equal, we explicitly define the elements in each set builder.

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A={1.3}

for event B

B={x|x is a number on a die}

The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as

B={1,2,3,4,5,6}

for event C

$C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1$

Hence the set c is C={1,3}

and for the set D {x| x is the number of heads when six coins re tossed }

In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6

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