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3 years ago

What is 69 x 287 pls help fast

Mathematics

2 answers:

Naily [24]3 years ago

7 0

Answer:

19803

Step-by-step explanation:

Multiplying the two numbers, you get 19803.

Rzqust [24]3 years ago

3 0

Answer:

The answer would be 19,803.

Step-by-step explanation:

What helps when doing multiplication is lining the problem up, like so:

287

x 69

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Find the exact area of the shaded region

GREYUIT [131]

Answer:

B: 192pi - 144

Step-by-step explanation:

Area of the sector = (120/360) * pi * r^2

Area of the sector = 1/3 * pi * 24 * 24

Area of the sector = 192 * pi

Now to find the area of the triangle.

The triangle is an isosceles triangle That means two of its sides are equal. They are equal to the radius of the circle, which is 24.

the small angles are equal to

2x + 120 = 180 Subtract 120 from both sides

2x = 60 Divide by 2

x = 60/2

x = 30

The height of the triangle is derived from sin(30) = opposite / hypotenuse

sin(30) = 1/2

hypotenuse = 24

1/2 = opposite / hypotenuse

1/2 = opposite / 24 Multiply both sides by 24

1/2 * 24 = opposite

opposite = 12

The height = 12

r = 24

Area of the triangle = 1/2 * 12 * 24

Area of the triangle = 144

So the area of the shaded area = 192*pi - 144 which looks like B

8 0

3 years ago

On a map, the distance between two towns is 2.5 inches, and 1 inch represents 6 miles. What is the actual distance between the t

White raven [17]

X=6.2*20
X=124 MILES BETWEEN THESE 2 TOWNS.
Hope this helps you:)

7 0

3 years ago

The sum of two numbers is 62, and their difference is 12. What are the numbers?

Ede4ka [16]

X + y = 62

x - y = 12

Move x over so y is alone in one of the above equations,

y = 62-x

Now substitte in y

x-(62-x) = 12

x - 62 + x = 12

2x = 12 + 62

2x/2 = 74/2

x = 37

Now for y sub in the x!

x + y = 62

(37) + y = 62

y = 62 - 37

y = 25.

So 25 + 37 =62

and 37 - 25 = 12

Method 2 :
Start by stacking the 2 original equations from above,

x+y=62
x-y=12
You want to "cancel" out one of the variable so,

x+y=62
+ + = +
x - y =12

So then you add,

2x = 74

2x / 2 = 74 / 2

x = 37.

Now you can once again sub it back into the original equation to solve for y,

x+y=62
(37)+y=62
y=62-37
y=25

Hope this helps!

5 0

2 years ago

Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

See attachment (1) for histogram

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

See attachment (2) for histogram

6 0

3 years ago

If (4,y) is an ordered pair of the function, then y =

irakobra [83]

that meen EVALUATE (y) at x=4

y=1

5 0

2 years ago