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3 years ago

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The height of the shortest man in history was 68 cm. This is 204 less than the height of the tallest man. how tall was the talle

st man?

1 answer:

Tanzania [10]3 years ago

4 0

This means the tallest man was 204 cm taller than the shortest. You add 204 to 68.

68 + 204 = 272

The tallest man was 272 cm tall.

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Regular hexagon ABCDEF is inscribed in a circle with center H. What is the image of segment BC after 120 degree clockwise rotati

fiasKO [112]

Answer: The answer would be EF

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3 years ago

In a recent survey, 73% of the community favored building a police substation in their neighborhood. If 14 citizens are chosen,

Archy [21]

Answer:

So, the probability is P=0.094.

Step-by-step explanation:

We know that in a recent survey, 73% of the community favored building a police substation in their neighborhood.

We get that p=0.73 and q=1-0.73=0.27.

We calculate the probability that exactly 8 of them favor the building of the police substation, if 14 citizens are chosen.

We get that n=14 and k=8.

We calculate the probability:

$P=C_k^n\cdot p^k\cdot q^{n-k}\\\\P=C_8^{14}\cdot 0.73^8\cdot 0.27^6\\\\P=3003\cdot 0.00003124\\\\P=0.094\\$

So, the probability is P=0.094.

4 0

3 years ago

Write a real world problem for 5/6 divided by 1/12. Use a model to solve.

sweet [91]

In the year of 2016 1000 high school students graduated from L.D. Bell, 5/6 of the students were A-Honor Roll. 1/12 of those were in NHS. How many of the student were in NHS out of the A-Honor Roll students?

6 0

3 years ago

Please solve! Thx! 6 = 20 - 3x

kumpel [21]

Answer:

5

Step-by-step explanation:

6=20-3x

combine

6-20=-3x

-14 = -3x

divide by -3

-14/-3 = -3x/-3

x = 4.666 repeatedly

x = 5 approximately

5 0

3 years ago

Andru [333]

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

◉ $\large\bm{ -4}$

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

Before performing any calculation it's good to recall a few properties of integrals:

$\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}$

$\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }$

So we apply the first property in the first expression given by the question:

$\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}$

And we solve the second integral:

$\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }$

Then we take the last equation and we subtract 10 from both sides:

$\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}$

And we divide both sides by 2:

$\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}$

Then we apply the second property to this integral:

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}$

Then we use the other equality in the question and we get:

$\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}$

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}$

We substract 8 from both sides:

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}$

• $\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}$

7 0

2 years ago