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kotykmax [81]
3 years ago
11

HELP ASAP ILL FRIEND U

Mathematics
1 answer:
Andrei [34K]3 years ago
8 0

Answer:

Undefined

Step-by-step explanation:

I did the quiz.

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Mary needs to buy 44 cookies for her party. If 6 cookies come in a package, how many packages of cookies does she buy?
xenn [34]
C because 6 times 8 is 48 which is greater than 42
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Step-by-step explanation:

4 0
3 years ago
If f(x) = 16x-30 and g(x) = 14(x)-6 for which value of x does (f-g) (x) = 0
almond37 [142]
This is a system where (f-g)(x) is wanted.
(16x-30)-(14x-6) .... 2x-24 *be careful with signs and parenthesis placement.

Now set 2x-24 = 0 and so x = 12
6 0
3 years ago
A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of funds' returns shows an averag
Feliz [49]

A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of 60 funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is

A. 0.037773 to 0.202227

B. 3.7773% to 20.2227%

C. 59.98786% to 61.01214%

D. 51.7773% to 68.2227%

E. 10.988% to 13.012%

Answer: E. 10.988% to 13.012%

Step-by-step explanation:

Given;

Mean x= 12%

Standard deviation r = 4%

Number of samples tested n = 60

Confidence interval is 95%

Z' = t(0.025)= 1.96

Confidence interval = x +/- Z'(r/√n)

= 12% +/- 1.96(4%/√60)

= 12% +/- 0.01214%

Confidence interval= (10.988% to 13.012%)

3 0
3 years ago
If log2 5 = k, determine an expression for log32 5 in terms of k.
lukranit [14]

Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

6 0
3 years ago
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