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3 years ago

How do you get the number 43 with the numbers 2,3,6,8 with using all numbers and only using them once.

Mathematics

1 answer:

iren [92.7K]3 years ago

8 0

Answer:

Multiply 8x6 and subtract 2 and 3

Step-by-step explanation:

8x6=48

48-3=45

45-2=43

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I didn't understand the question.

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2 years ago

A survey of 700 adults from a certain region asked, "What do you buy from your mobile device?" The results indicated that 59%

Kryger [21]

Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes $n_{1} = 400 and n_{2} = 300$

Proportion of mean $p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52$

Null hypothesis H0 : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

Alternative hypothesis H1:- p1 ≠ p2

a) The test statistic is

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }$

where $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

on calculation we get p = 0.56

now q =1-p = 1-0.56=0.44

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }\\ =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300} }$

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

conclusion:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) 95% confidence intervals

The confidence intervals are P± 1.96(√PQ/n)

we know that = $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of confidence intervals (0.523 ,0.596)

3 0

3 years ago

A company provides a monthly pension to its employees. A person retiring at 62 retires and receives a full monthly pension. If t

IRISSAK [1]

Answer:

The reduction is 8.6%

Explanation:

Call F the full monthly pension of a person retiring at 62.

If a person continues to work the pension grows at a rate of 6% per year, compounded monthly, so use the compounded growing formula:

$Pension=F(1+r/12)^{12t}$

Where r = 6 / 100 = 0.06, and t = number of years after retirement.

For retirement at 65.5:

t = 65.5 - 62 = 3.5

$Pension=F(1+0.06/12)^{12\times 3.5}=1.233F$

For retirement at 67:

t = 67 - 62 = 5

$Pension=F(1+0.06/12)^{12\times 5}=1.349F$

Percent reduction of people who retire at 65.5 compared to what they would receive at 67:

$(1.349F-1.233F)\times 100/(1.349F)=8.6\%$

8 0

3 years ago