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3 years ago

Solve 2(x + 3) = x-4

Mathematics

1 answer:

rusak2 [61]3 years ago

3 0

Answer:

x = -10

Step-by-step explanation:

$2(x + 3) = x - 4 \\ 2x + 6 = x - 4 \\ 2x - x = - 4 - 6 \\ x = - 10$

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When it is 11 a.m. in Johannesburg, it is 6 p.m. the same day in Tokyo, Japan. A flight leaves OR Tambo in Johannesburg at 6 a.m

kiruha [24]

Answer:

Dear User,

Answer to your query is provided below

The time will be 8am on Wednesday when flight arrive in Tokyo.

Step-by-step explanation:

Here, You can see that when 11 a.m. in Johannesburg, it is 6 p.m. the same day in Tokyo, Japan ; this means 7Hrs. gap.

Further, A flight leaves Johannesburg at 6 a.m. on Tuesday for Tokyo takes 18 hours in the air, with an additional 1 hour stop-over on land.

So, 6a.m. + 18hr.+1hr.+7hr. = 8am on Wednesday in Tokyo (Arrival)

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3 years ago

What is 77/8 in simplest form

julsineya [31]

Find a common multiplier then divide ex. 21/27 = 7/9

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3 years ago

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BartSMP [9]

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2 years ago

WHAT IS THE VOLUME OF THE PORTION OF THE AQUARIUM WITH WATER IN IT?

tensa zangetsu [6.8K]

The answer would be 32,200 50*28*23

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3 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