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3 years ago

5

There are 20 students in the community service club. The club needs 4 officers: a president, a vice president, a secretary, and

a treasurer.
Which expression can you use to find the number of ways to choose the officers from all of the members?

2 answers:

tiny-mole [99]3 years ago

6 0

well they could make a test they have to past for those who want to e officers.

steposvetlana [31]3 years ago

5 0

Answer:20! Over (20-4)!

Step-by-step explanation:

i got it correct

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A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1960 mailboxes t

MaRussiya [10]

Answer:

7088 m²

Step-by-step explanation:

The area of the half-cylinder is half the area of a cylinder with the same dimensions. This half-cylinder has a radius of 0.3 m and a length of 0.8 m, so its area is ...

A = πr(r +l) = π(0.3 m)(0.3 m+0.8 m) = 0.33π m² ≈ 1.0362 m²

The area of the bottom part of the box is the sum of its lateral area and its base area.

lateral area = Ph = (2(0.6m +0.8m))(0.75 m) = 2.1 m²

base area = lw = (0.8 m)(0.6 m) = 0.48 m²

The total area is then ...

total area = half-cylinder area + lateral area + base area

= 1.0362 m² +2.1 m² +0.48 m²

= 3.6162 m²

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The area for 1960 mailboxes will be 1960 times this area, or ...

area for week's production = 1960 × 3.6162 m² = 7087.752 m²

It will take about 7088 square meters of aluminum to make these mailboxes.

5 0

3 years ago

41 ÷ 533 long division r what

goblinko [34]

Answer:

13

Step-by-step explanation:

533/41

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2 years ago

If you invest $4000 at 9% interest compounded annually, in how many years will you have $20,000. Give your answer the nearest te

Masteriza [31]

Answer:

18.7 years

Step-by-step explanation:

This is a compound interest problem and the following variables have been given;

Principal = 4000; this is the amount o be invested

APR = 9%; this is the compound interest to be earned

Accumulated amount = 20,000

We are required to determine the duration in years. We apply the compound interest formula;

$A=P(1+r)^{n}$

$20000=4000(1+\frac{9}{100})^{n}\\20000=4000(1.09)^{n}\\5=(1.09)^{n}$

The next step is to introduce natural logarithms in order to determine n;

$ln5=nln(1.09)\\n=\frac{ln5}{ln(1.09)}\\n= 18.675$

The number of years required is thus 18.7 years

6 0

3 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

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

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

(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 <u>Two-sample t test statistics</u> 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, <u>test statistics</u> = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 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 <u>we fail to reject our null hypothesis</u>.

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 = <u>0.0662</u>

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2 years ago

the angle of depression from the top ski lift is 57 degrees if the vertical distance from the top and bottom ot the lift is 1200

dedylja [7]

Answer:

1200

Step-by-step explanation:

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3 years ago