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Aleksandr [31]
3 years ago
9

Assume every 10-digit natural number is a possible telephone number except those that begin with 0, 1 or 2. What fraction of tel

ephone numbers begins with 8 and ends with 8?
Mathematics
1 answer:
lana [24]3 years ago
5 0

Answer: 1/70

Step-by-step explanation:

This is a question that can also be interpreted as what is the probability of having the first number of a phone number to be 8 and the last number of the phone number to also be 8. This answer gives the fraction of the phone numbers that starts with 8 and end with 8.

Since three numbers (0,1,2) cannot start a phone number and we are left to pick from 7 numbers,

then the probability of figure "8" starting phone number = 1/7

Since all 10 numbers can possibly end a phone number,

then the probability of having figure "8" as the last digit of a phone number = 1/10

Hence probability of having "8" as the first and last digit of a phone number = fraction of total telephone numbers that begin with digit 8 and end with digit 8 = 1/7 × 1/10 = 1/70.

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A parcel weighs two thousand three hundred and forty grams. An item weighing seven hundred grams is removed from the parcel. Wha
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This is fairly simple, you just have to know how to set up an equation like this. It says that the package weighs 2340 grams. The item removed weights 700 grams. You just have to subtract these values to get the new weight of the package. 2340-700=1640
5 0
3 years ago
Sarah bought pencils and markers for school. She bought 3 times as many pencils as markers.
Serhud [2]

Answer:

Sarah bought 30 pencils

Step-by-step explanation:

1. Let define p, as the number of pencils bought, and m, as the number of markers bought.

2.

0.15p+0.30m=7.50

p=3m

3. Substitute p in the first equation to 3m since p=3m as defined.

0.15(3m) + 0.30m=7.50\\0.45m+0.30m=7.50\\0.75m=7.50\\m=10

plug m into equation p=3m to find the number of pencils bought

p=3m=3(10)=30

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B28%7D%7B54%7D%20%3D%20%5Cfrac%7B2x%2B8%7D%7B5x-4%7D" id="TexFormula1" title="\frac{2
tia_tia [17]

Step-by-step explanation:

step 1. what is the question? solve for x? okay.

step 2. 28/54 = (2x+ 8)/(5x - 4)

step 3. 28(5x - 4) = 54(2x + 8) (cross multiplication)

step 4. 14(5x - 4) = 27(2x + 8) (divide both sides by 2)

step 5. 70x - 56 = 54x + 216 (distributive property)

step 6. 16x = 272 (group x's and add 56 to both sides)

step 7. x = 17 (divide both sides by 16).

5 0
3 years ago
The figure is made up of a cylinder and a hemisphere. To the nearest whole number, what is the approximate volume of this figure
Novosadov [1.4K]

Hello!

The figure is made up of a cylinder and a hemisphere. To the nearest whole number, what is the approximate volume of this figure?  

Use 3.14 to approximate π .  


Enter your answer in the box.  in.³

Data: (Cylinder)

h (height) = 7 in

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

Adopting: \pi \approx 3.14

V (volume) = ?

Solving: (Cylinder volume)

V = \pi *r^2*h

V = 3.14 *2.5^2*7

V = 3.14*6.25*7

V = 137.375 \to \boxed{V_{cylinder} \approx 137.38\:in^3}

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

Adopting: \pi \approx 3.14

If: We know that the volume of a sphere is V = 4* \pi * \dfrac{r^3}{3} , but we have a hemisphere, so the formula will be half the volume of the hemisphere V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}

Formula: (Volume of the hemisphere)

V = 2* \pi * \dfrac{r^3}{3}

Solving:

V = 2* \pi * \dfrac{r^3}{3}

V = 2*3.14 * \dfrac{2.5^3}{3}

V = 2*3.14 * \dfrac{15.625}{3}

V = \dfrac{98.125}{3}

\boxed{ V_{hemisphere} \approx 32.70\:in^3}

Now, to find the total volume of the figure, add the values: (cylinder volume + hemisphere volume)

Volume of the figure = cylinder volume + hemisphere volume

Volume of the figure = 137.38 in³ + 32.70 in³

Volume\:of\:the\:figure =170.08 \to \boxed{\boxed{\boxed{Volume\:of\:the\:figure = 170\:in^3}}}\end{array}}\qquad\quad\checkmark

_______________________

I Hope this helps, greetings ... Dexteright02! =)

5 0
3 years ago
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