ASAP ASAP ASAP ASAP...

Vesna [10]

2 triangles with a base of 8 and height of 6 divide the base and height in half to get the Base and height of 1 triangle and then use 1/2 B•H to find the area of one triangle 1/2 (4•3) = 6

7 0

3 years ago

"A cash drawer contains 29 bills, all twenties and fifties. If the total value of the bills is $820, how many twenties and how m

maks197457 [2]

Answer:

There are 21 twenties and 8 fifties in the drawer.

Step-by-step explanation:

This question can be solved by a system of equations.

I am going to say that:

x is the number of twenties

y is the number of fifties.

A cash drawer contains 29 bills.

This means that x + y = 29.

The total value of the bills is $820.

This means that 20x + 50y = 820.

So

x + y = 29

20x + 50y = 820

From the first equation, x = 29 - y. So:

20x + 50y = 820

20(29 - y) + 50y = 820

580 - 20y + 50y = 820

30y = 240

y = 8

And

x = 29 - y = 29 - 8 = 21

There are 21 twenties and 8 fifties.

3 0

4 years ago

Read 2 more answers

A school choir needs to make T-shirts for its 75 members and has set aside some money in their budget to pay for them. The membe

Katyanochek1 [597]

50 + 3(B) + 50 + 3(R) + 50 + 3(Y) would be the equation.

7 0

3 years ago

A certain railway company claims that its trains run late 5 minutes on the average. The actual times (minutes) that 10 randomly

iragen [17]

Answer:

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 5 minutes

Sample mean, $\bar{x}$ = 9.130 minutes

Sample size, n = 10

Alpha, α = 0.01

Sample standard deviation, s = 1.4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 5\text{ minutes}\\H_A: \mu \neq 5\text{ minutes}$

We have to construct 99% confidence interval.

99% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.01} = \pm 3.249$

$9.130 \pm 3.249(\dfrac{1.4}{\sqrt{10}} ) = 9.130 \pm 1.4383 = (7.6917,10.5683)$

99% Confidence interval: (7.6917,10.5683)

Since the population mean does not lie in he calculated confidence interval, thus, we fail to accept the null hypothesis and accept the alternate hypothesis.

Option C) We would reject the claim that the trains run late an average of 5 minutes and conclude that they run late an average of more than 5 minutes since the 99% confidence interval exceeds 5.

6 0

3 years ago

Simplify the expression |12-5|-(6/2+1)^2

Zielflug [23.3K]

Answer:

-9

Alternative form:

- $3^{2}$

----------------------------------------------------

Steps:

1- Cross out the common factor

2- Calculate the sum or difference

3- Calculate the absolute value

4- Calculate the power

5- Calculate the sum or difference

5 0

2 years ago