Ask question

Ask question

All categories

3 years ago

11

Mrs. Rodriguez is going to use 6 1 3 material to make two dresses. The larger dress requires 3 2 3 yards of material. How much m

aterial will Mrs. Rodriguez have left to use on the smaller dress?

1 answer:

MArishka [77]3 years ago

5 0

So she need to use 500

You might be interested in

Heather wants to make a 36% acid solution.

bonufazy [111]

Answer:

.72x + .27 = .36(x + 3)

4 0

3 years ago

A tank can be filled by one pump in 3.2 hours and by another pump in 80 minutes. A third pump can drain the tank in 2 hours and

Mademuasel [1]

Answer:

Rounding to the nearest minute, it would take 95 minutes, or 1 hour and 35 minutes.

Step-by-step explanation:

Let's convert each time to minutes:

3.2 hours = 192 minutes

80 minutes = 80 minutes

2 hr 20 min = 140 minutes

Next, let's find the least common multiple:

LCM(192, 80, 140) = 6720

So let's say the volume of the tank is 6720 units. The speed of each pump is therefore:

Pump 1 = 6720 units / 192 minutes = 35 units/minute

Pump 2 = 6720 units / 80 minutes = 84 units/minute

Pump 3 = -6720 units / 140 minutes = -48 units/minute

Their combined speed is:

35 + 84 − 48 = 71 units/minute

So the time to fill the tank is:

6720 units / (71 units/minute) = 94.65 minutes

Rounding to the nearest minute, it would take 95 minutes, or 1 hour and 35 minutes.

6 0

3 years ago

Whats $4.23 divided by 50%

Bezzdna [24]

Here you go 4.32/ 0.50=8.64

3 0

3 years ago

Read 2 more answers

Find an explicit rule for the nth term of a geometric sequence where the second and fifth terms are -6 and 162, respectively. Pl

kicyunya [14]

Hello,

$u_{0} =a\\
 u_{1} =a*r\\
 u_{2} =a*r^{2}=-6\\
 u_{3} =a*r^{3}\\
 u_{4} =a*r^{4}\\
 u_{5} =a*r^{5}=162\\
....\\
\boxed{ u_{n} =a*r^{n}} \\

\dfrac{ 162}{-6} = \dfrac{ u_{5} }{ u_{3}} = \dfrac{ a*r^{5}}{ a*r^{2}} =r^3= -27\\
==\ \textgreater \ r=-3\\

 u_{2} =a*r^{2}=-6=a*(-3)^2 \ ==\ \textgreater \ a=-\frac{6}{9} =-\frac{2}{3}\\


\boxed{ u_{n} =-\frac{2}{3}*(-3)^{n}=(-1)^{n+1}*2*3^{n-1}} \\

$

8 0

3 years ago

Read 2 more answers

Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let AAA be the event that the six-sided d

myrzilka [38]

The question is incomplete. Here is the complete question.

Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let A be the event that the six-sided die is an even number and B be the event that the four-sided die is an odd number. Using the sample space of possible outcomes below, answer each of the following questions.

What is P(A), the probabillity that the six-sided die is an even number?

What is P(B), the probability of the four-sided die is an odd number?

What is P(A and B), the probability that the six-sided die is an even number <em>and</em> the four-sided die is an odd number?

Are events A and B independent?

a) Yes, events A and B are independent events.

b) No, events A and B are not independent events.

Answer: P(A) = 1/2

P(B) = 1/2

P(A and B) = 1/4

a) Yes, events A and B are independent events.

Step-by-step explanation: The event A is related to a six-sided die, so total possibilities is 6.

For a six-sided die to show a even number, there are 3 possibilities: (2,4,6)

so, P(A) = 3/6 = 1/2

The event B is for a 4-sided die, i.e. total possibilities is 4.

To show an odd number, there are 2 possibilities: (1,3).

Then, P(B) = 2/4 = 1/2

Now, the probability of occuring A and B is:

P(A and B) = P(A).P(B)

P(A and B) = 1/2*1/2

P(A and B) = 1/4

The events are <u>independent</u> events because the probability of A happening does not influence the occuring of event B.

3 0

4 years ago