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tatuchka [14]
3 years ago
9

How many different committees can be formed from 10 teachers and 38 students if the committee consists of 3 teachers and 2 ​stud

ents?
Mathematics
1 answer:
Marat540 [252]3 years ago
3 0
3 committees
Because there’s only enough teachers for the students to have 3 commiteees
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Solve the following problem. Round to one decimal place if necessary.
Luda [366]

Answer:

Sinθ = Opposite/hypotenuse

\sin(21)  =  \frac{x}{22}  \\ 0.358 =  \frac{x}{22}  \\ x = 22 \times 0.358 \\ x = 7.88

I hope I helped you^_^

7 0
3 years ago
Assume a recent sociological report states that university students drink 5.10 alcoholic drinks per week on average, with a stan
Lerok [7]

Answer:

Mean of sampling distribution = 5.10 alcoholic drinks per week

Standard deviation of the sampling distribution = 0.11

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 5.10 alcoholic drinks per week

Standard Deviation, σ = 1.3401

Sample size, n = 150

a) Mean of sampling distribution

The best approximator for the mean of the sampling distribution is the population mean itself.

Thus, we can write:

\mu_{\bar{x}} = \mu = 5.10

b) Standard deviation of the sampling distribution

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{1.3401}{\sqrt{150}} = 0.11

3 0
3 years ago
Anita brings 6 dolls to her grandma's house. These dolls represents 20% of Anita doll collection,as shown in the diagram.
8090 [49]

20/100 =x/6  

20 multiply 6 =120

120 divide by 100 =1.2

6 0
3 years ago
5. ¿A qué fracción es equivalente el número 27
nika2105 [10]

Answer:

Step-by-step explanation:

8 0
3 years ago
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Minchanka [31]
<h2><u>Answer with explanation</u>:</h2>

Let \mu be the distance traveled by deluxe tire .

As per given , we have

Null hypothesis : H_0 : \mu \geq50000

Alternative hypothesis : H_a : \mu

Since H_a is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.

Test statistic : z=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

where, n= sample size

\overline{x}= sample mean

\mu= Population mean

s=sample standard deviation

For n= 31,\ \sigma=8000,\ \overline{x}=46,800\ \&\ s=46,800, we have

z=\dfrac{46800-50000}{\dfrac{8000}{\sqrt{31}}}=-2.23

By using z-value table,

P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23)   [∵P(Z≤-z)=1-P(Z≤z)]

=1-0.9871=0.0129

 Decision : Since p value (0.0129) < significance level  (0.05), so we reject the null hypothesis .

[We reject the null hypothesis when p-value is less than the significance level .]

Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.

4 0
3 years ago
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