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Daniel [21]
3 years ago
15

Sam gave jen 1/2 of his jujubes. Jen ate 1/2 of the jujubes and gave the rest to klye. Kyle kept 8 of the jujubes and gave the l

ast 10 to Kim. How many jujubes did Jen eat?
Mathematics
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

Jen ate 18 jujubes

Step-by-step explanation:

Start with Kim. She was given 10. There is no remainder.

Kyle kept 8 from the total he was given, so Kyle had 18.

Let the number of Jujubes given to Jen = x

(1/2) x = 18    Multiply both sides by 2

2(1/2)x = 18*2

x = 36

Jen ate 1/2 of 36 which was 18.

Though you are not asked, Sam must have begun this with 2*36 = 72 Jujubes.


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Her insurance premiums are 150 per month.....there is 12 months in a year....so in 1 year, her premiums totaled : (150 * 12) = 1800

she paid a $ 500 deductible

she paid 20% of $ 3000 for the accident..
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7 0
3 years ago
Each person in random samples of 207 male and 253 female working adults living in a certain town in Canada was asked how long, i
Sergeu [11.5K]

Answer:

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females (P-value=0.155).

Step-by-step explanation:

The data is:

Males n1=207 M1=31.6 s1=24.0

Females n2=253 M2=29.3 s2=24.3

This is a hypothesis test for the difference between populations means.

The claim is that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.

Then, the null and alternative hypothesis are:

H_0: \mu_m-\mu_f=0\\\\H_a:\mu_m-\mu_f> 0

The significance level is 0.05.

The sample 1 (males), of size n1=207 has a mean of 31.6 and a standard deviation of 24.

The sample 2 (females), of size n2=253 has a mean of 29.3 and a standard deviation of 24.3.

The difference between sample means is Md=2.3.

M_d=M_m-M_f=31.6-29.3=2.3

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{24^2}{207}+\dfrac{24.3^2}{253}}\\\\\\s_{M_d}=\sqrt{2.783+2.334}=\sqrt{5.117}=2.262

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_m-\mu_f)}{s_{M_d}}=\dfrac{2.3-0}{2.262}=\dfrac{2.3}{2.262}=1.017

The degrees of freedom for this test are:

df=n_1+n_2-1=207+253-2=458

This test is a right-tailed test, with 458 degrees of freedom and t=1.017, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>1.017)=0.155

As the P-value (0.155) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.

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