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Karo-lina-s [1.5K]

3 years ago

13

If x=8 and y=14. Find y when x=48

1 answer:

Sladkaya [172]3 years ago

6 0

Answer:

Step-by-step explanation:

In picture shown.

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What does the combine place value strategy make it easier to find the difference

azamat

This may help you but I'm not so sure tho

7 0

3 years ago

Consider a manufacturing process with a quality inspection station. In the past, 10% of parts are defective. As soon as one defe

Kruka [31]

Answer:

0.9

Step-by-step explanation:

10% is equal to 0.1

The probability of having defective parts in a pile of parts is 0.1

Before the process is stopped, 1 part has to be defective.

In a pile of 9 parts, the probability that a part is defective 0.1 of 9, which is = 0.9 hence, approximately one (1) part will be defective in a pile of 9 parts and the process will be stopped.

Since there was no defective part among the first 6 parts, P(d) was 0

That is, probability of a defective part was zero.

7 0

3 years ago

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

<u>iii) Add the volume of the two frustrums</u>

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

Approximate the budget of #358879461.97 to the nearest million

diamong [38]

Answer:

359,000,000

Step-by-step explanation:

Brainliest plz :3

6 0

3 years ago

Andre says that 5 3/4 + 2 1/4 = 7 1/2 because 7 4/8 = 7 1/2. Identify his mistake. Draw a picture to prove that he is wrong.

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

The given expression is

$5\frac{3}{4}+2\frac{1}{4} =7\frac{4}{8}=7\frac{1}{2}$

We will rewrite the expression

$\frac{23}{4}+\frac{9}{4} =\frac{60}{8}=7\frac{1}{2}$

When we add the fraction

$\frac{23+9}{4}$ = $\frac{32}{4}$ = 8

While Andre says the answer is $7\frac{1}{2}$

because Andre added the denominator where he has to make a common denominator that should be 4 instead of 8.

Andre made mistake in this step

Picture is given below :

3 0

3 years ago