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3 years ago

Karen is seven years older than suzanne.Four years ago,karen was twice as old as suzanne was.How are they now?

2 answers:

7 0

Karen age=k
suzanne age=s

karen (k) is (=) 7 years older (+7) than suzanne(s)
k=7+s

four year ago means
s-4 and k-4

four years ago (-4 to all) karen (k) was (=) twice (2 times) suzzane (s)
k-4=2(s-4)
distribute
k-4=2s-8
add 4 to both sides
k=2s-4
we know that k=7+s so subsitute 7+s for k
7+s=2s-4
add 4 to both sides
11+s=2s
subtract s from both sides
11=s

suzanne=11
subsitute 11 for s in k=s+7

k=7+11
k=18

karen=18
suzanne=11

EleoNora [17]3 years ago

3 0

$K-Karen\ age\\S-Suzanne\ age\\K-4-Karen\ age\ 4\ years\ ago\\S-4-Suzanne\ age\ 4\ years\ ago\\\\ \left\{\begin{array}{ccc}K=S+7\\K-4=2(S-4)\end{array}\right\\\left\{\begin{array}{ccc}K=S+7&(1)\\K-4=2S-8&(2)\end{array}\right\\\\subtitute\ (1)\ to\ (2):\\\\(S+7)-4=2S-8\\S+7-4=2S-8\\S+3=2S-8\ \ \ \ |subtract\ 3\ from\ both\ sides\\S=2S-11\ \ \ |subtract\ 2S\ from\ both\ sides\\-S=-11\\\boxed{S=11}$

$Put\ the\ value\ of\ S\ to\ (1):\\\\K=11+7\\\boxed{K=18}\\\\\left\{\begin{array}{ccc}K=18\\S=11\end{array}\right\\\\Answer:\ Karen\ does\ 18\ years\ and\ Susanne\ does\ 11\ years.$

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Total cost = $48.10

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2 years ago

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Answer:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

Step-by-step explanation:

Data given and notation

n=75 represent the random sample taken

$\hat p=0.64$ estimated proportion of interest

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true proportion is higher than 0.5:

Null hypothesis: $p =0.5$

Alternative hypothesis: $p > 0.5$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

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3 years ago