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3 years ago

Please help me with 1-4

Mathematics

1 answer:

dsp733 years ago

8 0

For problem 1, it will be 25/100, which is equal to 25%
2) The relationship is that both, the bar model and the proportion, show that x is 1/4 of the actual amount. The bar graph is more a visual. However, in both cases, the x can be solved for to be 25%
3) The percent error for 23°C will be less since it is closer to 25°C, the actual amount. The percent error is greater for 5°C since it is farther for 25°C.

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How does probability apply to basketball?

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Depending on the player, a spectator can figure out the chance of the player making the basket :D

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Sumalee won 40 super bouncy balls playing horseshoes at her schools game last night. Later, she gave two to each of her friends.

Lina20 [59]

Before jumping into doing this problem, it is important to look at all the information's that have been made available in the question. Based on those given conditions the answer to the problem can be easily deduced.
Let us assume the number of friends Sumalee has = x
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2 years ago

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Can someone help?

babymother [125]

Answer:

0, 3, 8, 15, 24
y = x² + 2x

Step-by-step explanation:

The pattern is

Step 1 ⇒ 1*3 = 3
Step 2 ⇒ 2*4 = 8
Step 3 ⇒ 3*5 = 15

Next step would be

Step 4 ⇒ 4*6 = 24

Step zero would be

Step 0 ⇒ 0*2 = 0

The equation

y = x(x + 2) = x² + 2x

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2 years ago

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Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$

Then

$\dfrac{V_{\text{new flask}}}{V_{\text{flask}}} =\dfrac{8}{1}=8.$

Answer 2: correct choice is D.

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2 years ago

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Levart [38]

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2 years ago

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