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2 years ago

Which of these expressions is equivalent to log (20 27)

1 answer:

8 0

If the problem is referring to the equivalent logarithmic equation log (20 *27).
We can easily find and solve its equivalent expression using one of the many identities available in logarithmic.
We can have the expression:
log (20*27) = log 20 + log 27

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11. In a sample of n=6, five individuals all have scores of X=10 and the sixth person has a score of x=16. What is the mean for

ahrayia [7]

Answer:

Step-by-step explanation:

Given that in a sample of n=6, five individuals all have scores of X=10 and the sixth person has a score of x=16

i.e. the data entries are

10,10,10,10,10,16

Sum = 66

No of items n = 6

Average = sum/no of entries

= $\frac{66}{6} \\=11$

Average is 11.

mean of this sample is 11

8 0

3 years ago

Consider a uniform distribution from aequals4 to bequals29. (a) Find the probability that x lies between 7 and 27. (b) Find th

weeeeeb [17]

Answer:

a) 80% probability that x lies between 7 and 27.

b) 28% probability that x lies between 6 and 13.

c) 44% probability that x lies between 9 and 20.

d) 28% probability that x lies between 11 and 18.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value x between c and d, in which d is larger than c, is given by the following formula.

$P(c \leq x \leq d) = \frac{d - c}{b - a}$

Uniform distribution from a = 4 to b = 29

(a) Find the probability that x lies between 7 and 27.

So $c = 7, d = 27$

$P(7 \leq x \leq 27) = \frac{27 - 7}{29 - 4} = 0.8$

80% probability that x lies between 7 and 27.

(b) Find the probability that x lies between 6 and 13. 

So $c = 6, d = 13$

$P(6 \leq x \leq 13) = \frac{13 - 6}{29 - 4} = 0.28$

28% probability that x lies between 6 and 13.

(c) Find the probability that x lies between 9 and 20.

So $c = 9, d = 20$

$P(9 \leq x \leq 20) = \frac{20 - 9}{29 - 4} = 0.44$

44% probability that x lies between 9 and 20.

(d) Find the probability that x lies between 11 and 18.

So $c = 11, d = 18$

$P(11 \leq x \leq 18) = \frac{18 - 11}{29 - 4} = 0.28$

28% probability that x lies between 11 and 18.

3 0

2 years ago

160 bananas 20% decrease

ZanzabumX [31]

Inadequate problem statement. I'm assuming you meant "find the number of bananas lost in a 20% decrease from 160 bananas." That would be 0.20(160), or 32 bananas lost, with 128 bananas remaining. Please include all details when you share a problem statement.

3 0

3 years ago

Which is the correct answer

VashaNatasha [74]

Let s denote the number of pages in each issue of the sports magazine.

Let t denote the number of pages in each issue of the technology magazine.

Given that a sports magazine prints 12 issues per year and a technology magazine prints 10 issues per year.The total number of pages in all the issues of the sports magazine for one year is 32 more than the total number of pages in all the issues of the technology magazine for one year.

This can be written in equation as,

$12s=10t+32$

Since, it is given that, "Each issue of the sports magazine has 18 fewer pages than each issue of the technology magazine".

This can be written in equation as,

$s=t-18$

Thus, the set of equations that represent the given data is

$12s=10t+32$ and $s=t-18$

Hence, Option A is the correct answer.

7 0

3 years ago

Mrs. Zabel, the yearbook adviser, always likes to sell a large amount of yearbooks because after the initial set-up costs are pa

mart [117]

(50,45)(125,37.50)
slope(m) = (37.50 - 45) / (125 - 50) = -7.5/75 = -0.1

y = mx + b
slope(m) = -0.1
(50,45)...x = 50 and y = 45
now we sub and find b, the y int
45 = -0.1(50) + b
45 = - 5 + b
45 + 5 = b
50 = b

so ur equation is : y = -0.1x + 50

so it the staff sells 150 yearbooks...
y = -0.1x + 50
y = -0.1(150) + 50
y = -15 + 50
y = 35 <== price per yearbook

5 0

3 years ago