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1 year ago

A box contains

1 answer:

6 0

The probability of picking a hazelnut chocolate is 1/5 and the number of hazelnut chocolates in the box is 1

<h3>How to determine the probability</h3>

Note that the ratios = 5:4:2:3

toffee: coffee: orange: mint chocolates = 5:4:2:3:1

Total ratio = 15

Probability of picking a hazelnut chocolate = ratio of hazelnut chocolates/ total ratio

Probability of picking a hazelnut chocolate = $\frac{1}{15}$

The number of hazelnut chocolates in the box is gotten from the ratio which is 1

Therefore, the probability of picking a hazelnut chocolate is 1/5 and the number of hazelnut chocolates in the box is 1

Learn more about probability here:

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Which shows the decimals 34.8, 36.43, and 36.29 in order from greatest to least?

ruslelena [56]

All you do is add a zero on the end of this to make it to the nearest hundreth: 34.8(0)
That means that this is the correct order: 34.8, 36.43, 36.29 or Option B

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3 years ago

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Which expression is equivalent to 15 – 5x?

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Answer:

15 - 5x = 5 · 3 - 5 · x = 5(3 - x)

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3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

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3 years ago

Given the pdf of f(x) =2x for 0<x<2 and zero elsewhere, find the cdf

Reptile [31]

Hope it helps.........

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2 years ago

Element X decays radioactively with a half life of 14 minutes. If there are 760 grams of Element X, how long, to the nearest ten

Mkey [24]

Answer:

85.5 minutes

Step-by-step explanation:

The amount of an element that will remain after time t can be expressed as a function of initial amount N0, time t, and half life th as;

Nt = N0 × e^(-λt)

Where;

Decay constant λ = ln(2)/th, substituting into the equation;

Nt = N0 × e^(-ln(2)t/th)

We need to make t the subject of formula;

Nt/N0 = e^(-ln(2)t/th)

ln(Nt/N0) = -ln(2)t/th

t = ln(Nt/N0) ÷ -ln(2)/th

Given;

Initial amount N0 = 760g

Final amount Nt = 11 g

Half life th = 14 minutes

the nearest tenth of a minute, would it take the element to decay to 11 grams can be derived using the formula;

t = ln(Nt/N0) ÷ -ln(2)/th

Substituting the given values;

t = ln(11/760) ÷ -ln(2)/14

t = 85.5 minutes

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2 years ago