All categories

2 years ago

In a group of students, 30 played chess, 19 played volleyball, 25 played

Mathematics

1 answer:

saul85 [17]2 years ago

5 0

Answer:

Explained below.

Step-by-step explanation:

Denote the events as follows:

C = chess

V = volleyball

B = basketball

The data provided is as follows:

n (C) = 30

n (V) = 19

n (B) = 25

n (C ∩ V) = 14

n (B ∩ V) = 8

n (B ∩ C) = 15

n (C ∩ V ∩ B) = 5

Consider the Venn diagram below.

The number of students who played only chess is marked in pink:

n (Only C) = 6

The number of students who played only volleyball is marked in blue:

n (Only V) = 2

The number of students who played only basketball is marked in orange:

n (Only B) = 7

The number of students who played all three is marked in grey:

n (C ∩ V ∩ B) = 5

You might be interested in

Consider the first five steps of the derivation of The Quadratic function

Burka [1]

Answer:

Step-by-step explanation:

pick like terms

x² + b²/4a² = -c / a + b²/4a²

b²/4a² = (b/2a)²

x² + (b/2a)² = -c/a + (b/2a)²

(x + b/2a)² = -c/a + (b/2a)² = -c / a + b²/4a² = (-4ac+ b²)/4a²

(x + b/2a)² = (-4ac+ b²)/4a²

square root both sides

√{(x + b/2a)²} = √{(-4ac+ b²)/4a²}

x + b/2a = √(-4ac+ b²) / √(4a²) = √(-4ac+ b²) / 2a = √( b²-4ac) / 2a

x + b/2a = √( b²-4ac) / 2a

subtract b/2a from both sides

x + b/2a -b/2a = {√( b²-4ac) / 2a } -b/2a

x = -b/2a + {√( b²-4ac) / 2a }

the l.c.m is the same

x = {-b±√( b²-4ac)}/2a

7 0

2 years ago

"Express as a mixed numeral: 39/8" is a problem found in a fifth-grade arithmetic textbook. This problem is BEST solved through:

Scilla [17]

Answer:

c. an algorithm

Step-by-step explanation:

Most math problems are best solved using an algorithm. Certainly, converting numbers from one form to another is easily done that way.

_____

The integer part is the whole-number quotient of the division. The fractional part is the remainder divided by the denominator:

39/8 = 32/8 + 7/8 = 4 7/8

5 0

2 years ago

What can you say about the end behavior of the function f(x)=log10(5x-1)

natima [27]

The given function is
f(x) = log₁₀(5x-1)

As x -> -∞, the argument of the log function becomes a large negative number.
Because the log of a negative number is undefined, f(x) is undefined as x -> -∞.

As x -> +∞, the argument of the log function becomes a large positive number.
Therefore f(x) -> +∞ as x -> +∞.

Answer:
As x -> -∞, f(x) is undefined.
As x-> +∞, f(x) -> +∞.

6 0

2 years ago

Read 2 more answers

Which of these is a recurring cost of car ownership?

Vlad1618 [11]

Answer:

D. Windshield replacement

Step-by-step explanation:

A Recurring Cost is a regularly occurring cost or estimated cost which is documented with one record—a Recurring Cost record—that describes the income or expense and its pattern (how often it occurs, the rate at which it increases or decreases, the time period during which the cost applies, and so forth)

5 1

2 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 $2P(T$ 0 (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

2 years ago