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kupik [55]
3 years ago
14

How much will be left after 24 hrs if it’s half life is 6 hours?

Mathematics
1 answer:
Aleks [24]3 years ago
3 0
Hello there! A half-life is basically a decay of 50%. We will find the amount left after one day by exponential decay. The formula for exponential decay is A(1 - r)^t, where A = initial amount, r = rate (in decimal form), and t = time (or in this case, amount of half lives). The half-life is 6 hours and 24 goes into 6 four times. That's 4 half lives. Let's start by subtract 50% (0.5) from 1. 1 - 0.5 is 0.5. Now, we will use 0.5 and raise it to the 4th power, because the 4th power represents 4 half lives. 0.5^4 is 0.0625. Leave this number on the calculator. Do not round. Multiply that number by 448 to find the amount left. When you do, you get 28. There. 28 grams will be left after 24 hours.
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Step-by-step explanation:

(a - b)² = a² + 2ab + b²

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                                   = (2b - 1)²

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2 years ago
Researchers studying the acquisition of pronunciation often compare measurements made on the recorded speech of adults and child
dolphi86 [110]

Answer:

(A) The additional information that is needed to confirm about the conditions for this test have been met is ‘Population is approximately normal.

(B) The test statistic = 1.9965

(C1) P-value = 0.0556

(C2) There is no significant difference between mean VOT for children and adults.

D1) It is possible to make type II error.

(D2) There should be a difference in the VOT of adults and children.

Step-by-step explanation:

Let na be the number of adults = 20

xa mean VOT for adults = 88.17

sa standard deviatiation of VOT for adults = 24.74

Let nc be the number of children = 10

xc mean VOT for children =60.67

sc standard deviatiation of VOT for children = 39.89

(A) From the information the population variances are unknown and the two sample are assumed to be independent and the sample the sample size are smaller that is (n<30).

This indicates that the additional information that is required for the conditions of the test to be satisfied is 'distribution of the population'. the addition assumption to be made is, that the population distribution is normal.

(b) Calculating the test statistics using the formula;

t = (xa -xc)/SE - d

where SE = standard deviation , d= hypothesized difference = 0

But SE = √sa²/na +sc²/nc

           = √24.74 ²/20 + 39.89²/10

           = √189.72559

           = 13.774

Substituting into test statistics equation, we have

t = (xa -xc)/SE - d

  = (88.17 - 60.67/13.774

  = 1.9965

Therefore the test statistic is 1.9965

(c) Calculating the p-value, we have;

Degree of freedom = na + nc -2

                                 = 10+ 20 -2

                                 = 28

The p-value for t=1.9965 at 28 degrees of freedom and 0.05 level of significance is 0.0556.

The p-value 0.0556 is greater than given level of significance 0.05 hence we fail to reject the null hypothesis and conclude that there is no significant difference between mean VOT for children and adults.

(D1) From the information in part (C) the null hypothesis is not rejected.

Since the null hypothesis is not rejected, there might be a chance that not rejecting null hypothesis would be wrong. In this type of situation the error that can occur would be type II error.

(D2) The type of error describe in the context of this study is obtained by the concept of the type II error which tells that the null hypothesis is not rejected when it is actually false.

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3 years ago
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