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3 years ago

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Maria has 1.35.she has only quarters and nickels.draw two possible group of coins.maria could have.use 25s to to show quarters a

nd 5s to show nicjeks

1 answer:

Papessa [141]3 years ago

8 0

Go yo abswers.com and there ut would see tye answer it's very helpful

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Use formulas to find the lateral area and surface area of the prism.

TiliK225 [7]

Step-by-step explanation:

Finding surface area

To find the surface area of a prism, use the formula SA=2B+ph, where SA stands for surface area, B stands for area of the base of the prism, p stands for the perimeter of the base, and h stands for the height of the prism. Since this is a triangular prism, substitute the area formula of a triangle for B.

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2 years ago

Solve 5y - 2 = 3y + 14

Valentin [98]

Answer: y=8

Step-by-step explanation:

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3 years ago

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In a recent year, one United States dollar was equal to about 82 Japanese yen. How many Japanese yen are equal to $100? $1,000?

alexandr402 [8]

Well if $1=82 than $100 (which is 100 times 1) would equal 82·100=8,200
$1,000·82=82,000
$10,000·82=820,000 :) hope this helps!

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3 years ago

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Prove that the diagonals of a rectangle bisect each other.

Umnica [9.8K]

Answer:

2

Step-by-step explanation:

example A(2a,0),B(2b,0)

C(2b,2c),D(2a,2c)

mid point of AC=((2a+2b)/2,(0+2c)/2)=(a+b,c)

mid point of BD=((2b+2a)/2,(0+2c)/2)=(a+b,c)

∴midpoint of diagonals same or diagonals bisect each other.

7 0

3 years ago

a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou

Fed [463]

Answer:

$\displaystyle P(A)=\frac{6}{11}$

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

$\Omega=\{WW,WN,NW,NN\}$

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

$A=\{WN,NW\}$

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

$\displaystyle \frac{5}{11}$

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

$\displaystyle \frac{6}{10}$

Thus the option WN has the probability

$\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}$

Now for the second option NW. The initial probability to pick a non-white counter is

$\displaystyle \frac{6}{11}$

The probability to pick a white counter is

$\displaystyle \frac{5}{10}$

Thus the option NW has the probability

$\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}$

The total probability of event A is the sum of both

$\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}$

$\boxed{\displaystyle P(A)=\frac{6}{11}}$

7 0

3 years ago