Math question i kind of need this done quicker, however no rush..! :) x=-2y x-y=9

What is the area of this trapezoid?

dimulka [17.4K]

Answer:

80

Step-by-step explanation:

area of a trapezoid: (base1+base2)/2 * height

(3+17)/2 *8 = 10*8 = 80

6 0

3 years ago

Find the value of the following. (a) 180 % of 200g

Natasha_Volkova [10]

Answer:

$\frac{180| }{100} \times 200 = 360$

3 0

3 years ago

Read 2 more answers

Alan has twice as many dimes as he has quarters and 10 more nickels than dimes. The remaining 35 coins are pennies. How many of

zaharov [31]

Answer:

He has 3 quarters , 6 dimes and 16 nickles

Step-by-step explanation:

∵ Dimes is d , Quarter is q , nickels is n and pennies is p

∵ Quarter = 25 cents

∵ Dime = 10 cents

∵ Nickle = 5 cents

∵ Penny = 1 cent

∵ The remaining 35 coins are pennies ⇒ 35 cents

∵ He has $2.50 = 2.50 × 100 = 250 cents

∴ The value of quarters , dimes and nickels = 250 - 35 = 215 cents

∴ 25q + 10d + 5n = 215 ⇒ (1)

∵ d = 2q

∴ q = 1/2d ⇒ (2)

∵ n = 10 + d ⇒ (3)

Substitute (2) and (3) in (1)

∴ 25(1/2 d) + 10d + 5(10 + d) = 215

∴ 12.5d + 10d + 50 + 5d = 215

∴ 27.5d = 215 - 50

∴ 27.5d = 165

∴ d = 165 ÷ 27.5

∴ d = 6

∴ q = 1/2 (6) = 3

∴ n = 10 + 6 = 16

∴ He has 3 quarters , 6 dimes and 16 nickles

7 0

3 years ago

Read 2 more answers

CAN SOMEBODY PLEASE HELP ME!!!!!!!!!!

lana66690 [7]

The answers to the blanks are

1. 600,

2. 120,

3. 120,

4. 180.

Explanation:

The fence means perimeter around the court. So a tennis court's perimeter is 600 feet fence. The perimeter of a rectangle is 2 times the sum of the rectangle's length and the rectangle's width.
It is given that length equals 60 more than width i.e. l = w + 60, where l is the length of the court and w is the width of the court.
The perimeter of a court = 600 = 2 (l + w) = 2l + 2w = 2 (w +60) + 2w, this becomes, 2w + 120 + 2w = 600; 4w = 480, w = 120.
Since l = w + 60, l = 120 + 60 = 180. So length of a court is 180 feet and the width of a court is 120 feet.

6 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).

$P(Y=0) + P(Y \geq 1) = 1 \implies P(Y \geq 1) = 1 -P(Y=0) = 1 - 0.0821 = 0.918$

3 0

3 years ago