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guapka [62]
3 years ago
12

Find the value of the following expression (2^8 • 3^-5 • 6^0) write your answer in simplified form show all your steps Help Fast

Mathematics
1 answer:
vitfil [10]3 years ago
6 0
{( {2}^{8} \times {3}^{ - 5} \times {6}^{0} })^{ - 2} \\ = \frac{1}{ {( {2}^{8} \times {3}^{ - 5} \times {6}^{0} })^{2} } \\ = \frac{1}{ {(256 \times \frac{1}{ {3}^{5} } \times 1 })^{2} } \\ = \frac{1}{ {(256 \times \frac{1}{243} \times 1 })^{2} } \\ = \frac{1}{ ({ \frac{256}{243} })^{2} } \\ = \frac{1}{ \frac{ {256}^{2} }{ {243}^{2} } } \\ = \frac{1}{ \frac{65536}{59049} } \\ = \frac{59049}{65536}
The answer is 59049/65539
The value is 0.901
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Distributive property of 18+24
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Distributive property: <em>a(b + c) = ab + ac</em>.

We have 18 + 24

18 = 6 · 3

24 = 6 · 4

Therefore

18 + 24 = 6 · 3 + 6 · 4 = 6 · (3 + 4)

<h3>Answer: 18 + 24 = 6 · (3 + 4)</h3>
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A college student realized that he was spending too much money on music. For the remaining 5 months of the year his goal is to s
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70 dollars

Step-by-step explanation:

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3 years ago
Fwerwer34c43t45f345t45f3554t4f556gf5f5f6g65rtytwhat is -4+1/5
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3 years ago
-1/2 + 4/9
stira [4]

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4 0
2 years ago
.A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and
Leni [432]

Answer:

a) Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

b) t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236  

The degrees of freedom are given by:

df = n-1 = 80-1=79

The p value for this case taking in count the alternative hypothesis would be:

p_v =P(t_{79}>2.236)=0.0141  

Step-by-step explanation:

Information given

\bar X=130 represent the sample mean for the amount spent each shopper

s=40 represent the sample standard deviation

n=80 sample size  

\mu_o =120 represent the value to verify

t would represent the statistic    

p_v represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:  

Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

The statistic for this case would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Replacing the info given we got:

t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236  

The degrees of freedom are given by:

df = n-1 = 80-1=79

The p value for this case taking in count the alternative hypothesis would be:

p_v =P(t_{79}>2.236)=0.0141  

5 0
3 years ago
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