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Elina [12.6K]
3 years ago
9

In how many different ways can a man divide 7 different gifts among his 3 children if the eldest is to receive 3 gifts and the o

thers 2 each
Mathematics
1 answer:
krok68 [10]3 years ago
7 0

The order in which gifts are received doesn't matter - if child X gets toy 1 then toy 2, it's the same as giving child X toy 2 then toy 1 - so we are counting combinations.

The eldest child receives 3 of the 7 gifts, so they have

\dbinom 73 = \dfrac{7!}{3!(7-3)!} = 35

possible choices of gifts.

The next child receives 2 of the remaining 4 gifts, so they have

\dbinom 42 = \dfrac{4!}{2!(4-2)!} = 6

choices.

The last child receives the remaining 2 gifts, and there is only

\dbinom22 = \dfrac{2!}{2!(2-2)!} = 1

way to select the gifts for them.

By the multiplication using, the total number of ways of distributing 7 gifts among 3 children in the prescribed way is 35 • 6 • 1 = 210.

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Which expression is a factor of 12x2 +29x -8
Temka [501]

Answer:

Dear user,

Answer to your query is provided below

(4x−1)(3x+8) is a factor of 12x^2+29x-8

8 0
3 years ago
kurt spots a bird sitting at the top of a 40 foot tall telephone pole. If the angle of elevation from the ground where he is sta
Likurg_2 [28]

angle of elevation, aoe = 59 deg

height, h = 40ft

distance from pole, dfp

using

tan 59deg = h/dfp

=》 dfp = h/ tan 59deg = 40/1.67 = 23.95ft

3 0
3 years ago
Read 2 more answers
Hello again,
Nataly [62]

Answer: 395

Step-by-step explanation:

201+99+95

5 0
3 years ago
Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s
stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  

Then we have n_{1} = 25, \bar{x}_{1} = 2.04, \sigma_{1} = 0.010 and n_{2} = 20, \bar{x}_{2} = 2.07, \sigma_{2} = 0.015. The pooled estimate is given by  

\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative).  

The test statistic is T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}} and the observed value is t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257. T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value t_{0} falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by 2P(T0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}, i.e.,

-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}

where t_{0.025} is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

-0.03\pm(2.0167)(0.012459)(0.3), i.e.,

(-0.0225, -0.0375)

8 0
3 years ago
What is the 17th term in the arithmetic sequence in which a6 is 101 and a9 is 83
monitta

Answer:

The 17th term in arithmetic sequence is 68

Step-by-step explanation:

The general formula of arithmetic sequence is:

aₙ = a₁ + (n – 1)d.

We are given a₆ = 101 and a₉ = 83 and we need to find a₁₇

To find the term a₁₇ we should know a₁ and d. So we would find both

a₆ = a₁ +(6-1)d

101 = a₁ +(5)d

101 = a₁ +5d     eq(1)

and

a₉ = a₁ +(9-1)d    

83 = a₁ + 8d       eq(2)

Subtracting eq(2) from eq(1)

101 = a₁ +5d

83 = a₁ + 8d

-       -     -

__________

18 = -3d

=> d = 18/-3

=> d = -6

Putting value of d in eq(1)

101 = a₁ + 5d

101 = a₁ + 5(-3)

101 = a₁ -15

=> a₁ = 101+15

=> a₁ = 116

Now finding a₁₇:

aₙ = a₁ + (n – 1)d.

a₁₇ = 116 +(17-1)(-3)

a₁₇ = 116+(16)(-3)

a₁₇ = 116 - 48

a₁₇ = 68

So, the 17th term in arithmetic sequence is 68

3 0
3 years ago
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