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3 years ago

What are the multiples of 4

Mathematics

2 answers:

N76 [4]3 years ago

8 0

1 and 2 are the factors its a compatable number

mario62 [17]3 years ago

6 0

The multiples of 4 are 4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,100,104,108 and so on

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Find the t-value such that the area in the right tail is 0.2 with 5 degrees of freedom.

evablogger [386]

Answer:

0.920

Step-by-step explanation:

To calculate this, we proceed to the t-table

Using degree of freedom 5 and significance level of 0.2, the t-value is 0.920

3 0

3 years ago

The ratio of the measures of the sides of a triangle is 3:4:5. Its perimeter is 48 inches. Find the scale factor as a decimal, a

Trava [24]

Given:

The ratio of the measures of the sides of a triangle is 3:4:5.

Its perimeter is 48 inches.

To find:

The scale factor as a decimal, and the measure of each side of the triangle.

Solution:

Let x be the scale factor. Then the measures of sides of the triangle are 3x, 4x and 5x.

The perimeter of the triangle is 48 inches. It means the sum of all sides of the triangle is 48 inches.

$3x+4x+5x=48$

$12x=48$

Divide both sides by 12.

$x=\dfrac{48}{12}$

$x=4$

Now, the measures of sides are:

$3x=3(4)$

$3x=12$

Similarly,

$4x=4(4)$

$4x=16$

And,

$5x=5(4)$

$5x=20$

Therefore, the scale factor is 4 and the measures of sides are 12, 16 and 20.

5 0

2 years ago

Read 2 more answers

Two cars leave Phoenix and travel along roads 90 degrees apart. If Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mp

romanna [79]

Answer:

C. 210 miles

Step-by-step explanation:

We have been given that two cars leave Phoenix and travel along roads 90 degrees apart. Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mph.

We will use distance formula and Pythagoras theorem to solve our given problem.

$\text{Distance}=\text{Speed}\times \text{Time}$

$\text{Distance covered by Car 1 in 3.5 hours}=\frac{42\text{ Miles}}{\text{Hour}}\times \text{3.5 hour}$

$\text{Distance covered by Car 1 in 3.5 hours}=147\text{ Miles}$ .

Since car 1 leaves 30 minutes before car 2, so car 2 will travel for only 3 hours when car 1 will travel for 3.5 hours.

$\text{Distance covered by Car 2 in 3 hours}=\frac{50\text{ Miles}}{\text{Hour}}\times \text{3 hour}$

$\text{Distance covered by Car 2 in 3 hours}=150\text{ Miles}$

Since both car travel along roads 90 degree apart, therefore, the distance between both cars after Car 1 has traveled 3.5 hours would be hypotenuse with legs 147 and 150.

$\text{Distance between both cars}=\sqrt{147^2+150^2}$