BRAINLIESTTTT PLEASE ASAP!!!

Suppose you work for a life insurance company. At your job, you use the following table, which shows the number of people out of

irina [24]

Oh boi I'm gonna fail, I can't even geuss

3 0

2 years ago

Read 2 more answers

What does the initial value mean for this function?

Anit [1.1K]

She saves 10 per week.......after 4 weeks, she had 90

so if she started with nothing, after 4 weeks she would have 40.....but for her to have 90, she would have had to have started with (90 - 40) = 50.

Think about it....she starts with 50....and then saves 10 for 4 weeks giving her 40......and now she has 40 + 50 = 90

so ur answer is : Roberta had $ 50 before she started to save money each week

5 0

2 years ago

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

Let $\mu_1$ = mean number of times men order take-out dinners in a month.

$\mu_2$ = mean number of times women order take-out dinners in a month

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, test statistics = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = 0.0662

4 0

2 years ago

In ΔJKL, j = 93 cm, k = 82 cm and l=86 cm. Find the measure of ∠J to the nearest degree.

zhuklara [117]

Answer:

5

Step-by-step explanation:

Answer:

85t

Step-by-step explanation:

1. Volume of a cone: 1. V = (1/3)πr2h 2. Slant height of a cone: 1. s = √(r2 + h2) 3. Lateral surface area of a cone: 1. L = πrs = πr√(r2 + h2) 4. Base surface area of a cone (a circle): 1. B = πr2 5. Total surface area of a cone: 1. A = L + B = πrs + πr2 = πr(s + r) = πr(r + √(r2 + h2))

hope this helps

8 0

2 years ago

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Kevin bicycles a total of 2,340 miles last year. On average, about how many miles did he bike each week? 195 45 44 46

svlad2 [7]

In order to find the average you have to add up all the data values and then find the sum of the data values.

$195 + 45 + 44 + 46 = 330\\\frac{330}{4} = 82.5\\\\$

So on average he ran about 82.5 miles.

7 0

2 years ago