What could you call this shape? rhombus parallelogram trapezoid rectangle

can company makes a cylindrical can that has a radius of 6 cm and a height of 10 cm. One of the company's clients needs a cylind

shutvik [7]

the new radius be to meet the client's need is 4.9 cm .

Step-by-step explanation:

Here we have , can company makes a cylindrical can that has a radius of 6 cm and a height of 10 cm. One of the company's clients needs a cylindrical can that has the same volume but is 15 cm tall. We need to find What must the new radius be to meet the client's need . Let's find out:

Let we have two cylinders of volume $V_1 , V_2$ with parameters as follows :

$r_1=6cm\\h_1=10cm\\r_2=?\\h_2=15cm$

We know that volume of cylinder is $\pi r^2h$ , According to question volume of both cylinder is equal i.e

⇒ $V_1=V_2$

⇒ $\pi (r_1)^2h_1= \pi (r_2)^2h_2$

⇒ $(r_1)^2h_1= (r_2)^2h_2$

⇒ $\frac{(r_1)^2h_1}{h_2}= (r_2)^2$

⇒ $(r_2) =\sqrt{ \frac{(r_1)^2h_1}{h_2}}$ Putting all values

⇒ $(r_2) =\sqrt{ \frac{(6)^2(10)}{15}}$

⇒ $(r_2) =\sqrt{ \frac{36(10)}{15}}$

⇒ $(r_2) =\sqrt{ \frac{360}{15}}$

⇒ $(r_2) =\sqrt{24}$

⇒ $(r_2) =4.9cm$

Therefore , the new radius be to meet the client's need is 4.9 cm .

7 0

3 years ago

Read 2 more answers

Evaluate \dfrac32y-3+\dfrac53z 2 3 y−3+ 3 5 zstart fraction, 3, divided by, 2, end fraction, y, minus, 3, plus, start fracti

Arlecino [84]

Answer:

11

Step-by-step explanation:

Given:

3/2y - 3 + 5/3z

When

y=6

z=3

3/2y - 3 + 5/3z

Substitute the value of y and z

3/2(6) - 3 + 5/3(3)

=18/2 - 3 + 15/3

=9-3+5

=6+5

=11

4 0

3 years ago

Read 2 more answers

Which line shows the first error in the situation? 3(× -2)= 18 (1). 3x -6 =18 (2). 3x -6 + 6 = 18 + 6. (3) 3x =24. (4). x = 21.

dedylja [7]

C. line 4 is the first one
C. -15 for the second part

7 0

4 years ago

sandy has 16 roses 8 daisies and 32 tulips she wants to arrange all the flaowers in bouquets each bouquet has the same number of

VMariaS [17]

2 Daises, 4 Roses, and 8 Tulips

6 0

3 years ago

Read 2 more answers

The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable

topjm [15]

Answer:

(A) 0.15625

(B) 0.1875

(C) Can't be computed

Step-by-step explanation:

We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let X = Amount of time taken by student to complete a statistics quiz

So, X ~ U(32 , 64)

The PDF of uniform distribution is given by;

f(X) = $\frac{1}{b-a}$ , a < X < b where a = 32 and b = 64

The CDF of Uniform distribution is P(X <= x) = $\frac{x-a}{b-a}$

(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)

P(X > 59) = 1 - P(X <= 59) = 1 - $\frac{x-a}{b-a}$ = 1 - $\frac{59-32}{64-32}$ = $1-\frac{27}{32}$ = 0.15625

(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)

P(X <= 43) = $\frac{43-32}{64-32}$ = $\frac{11}{32}$ = 0.34375

P(X < 37) = $\frac{37-32}{64-32}$ = $\frac{5}{32}$ = 0.15625

P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

(C) Probability that student complete the quiz in exactly 44.74 minutes

= P(X = 44.74)

The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.

3 0

4 years ago