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3 years ago

10

a Can of soup has a diameter of 2.5 inches and a height of 4 inches. how much metal is needed to make the can? round your answe

r to the nearest tenth.

1 answer:

Ratling [72]3 years ago

7 0

First the 2 bases have surface area pi*1.25^2*2

Then the side

2.5pi*4=10pi

total of 9.817+31.415

=41.232

This approximation could be further round down for this problem to 41.2

Which makes choice C the correct choice.

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If 10% of a number is 44, 5% of the number is

inn [45]

Answer:

22

Step-by-step explanation:

I hope this helps <3

5 0

2 years ago

What are the measures of angles a,b, and c? Show your work and explain your answers.

Virty [35]

Answer:

c is 110 degrees

b is 70 degrees

a is 20 degrees

Step-by-step explanation:

for c, 180-70=110

for b, 180- angle c=180-110=70

for a, 180-90-angle b=180-90-70=20

8 0

2 years ago

Read 2 more answers

In the triangle below, ∠Q is a right angle. Which ratio would be equivalent to the ratio of sin∠R?

MrMuchimi

Answer:

b) cos∠S

Step-by-step explanation:

sin ∠R = SQ/SR

<u>This is same as:</u>

cos S

Correct option is b)

4 0

3 years ago

How many numbers greater than zero and less than 50 equal the product of two consecutive positive integers?

spin [16.1K]

Consecutive numbers would be like 2 and 3, or 7 and 8.

All we need to do is keep multiplying pairs of consecutive numbers until we get above 50.

1 × 2 = 2 (that's one.)
2 × 3 = 6 (two)
3 × 4 = 12 (three)
4 × 5 = 20 (four)
5 × 6 = 30 (five)
6 × 7 = 42 (six...)
<em>7 × 8 = 56 > 50</em>
We have a total of 6 numbers that equal the product of 2 consecutive intergers<em>

</em>

7 0

2 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

2 years ago