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2 years ago

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Mathematics

1 answer:

GaryK [48]2 years ago

5 0

Answer:

The computation of the angles show that the value of sin x and cos y will be 4/5 and 3/5 respectively.

How to calculate the angles?

From the information, we've to compute the length of the hypothenuse using Pythagoras theorem.

Therefore, 4² + 3² = hypothenuse²

hypothenuse² = 16 + 9

hypothenuse² = 25

hypothenuse = ✓25.

hypothenuse = 5

Therefore, sin x = opposite/hypothenuse = 4/5

cos y = adjacent/hypothenuse = 3/5

In conclusion, the values are 4/5 and 3/5.

Step-by-step explanation:

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How many solutions does the system have? y=−2x+4

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There is only one solution and the solution is (0,4).

Step-by-step explanation:

The given system has equations;

$y=-2x+4$

and

$y=x^2-2x+4$

We equate the two equations to determine their point of intersection;

$x^2-2x+4=-2x+4$

$\Rightarrow x^2-2x+2x+4-4=0$

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There is only one solution and the solution is (0,4).

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You have 6 reindeer, Rudy, Jebediah, Ezekiel, Lancer, Gloopin, and Balthazar, and you want to have 5 fly your sleigh. You always

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Answer:

Step-by-step Answer:

6 reindeer, from which we fly 5, in N1 ways.

Each of the five must be arranged in N2 ways.

The total number of arrangements is therefore N1*N2 arranglements.

Note: C(n,r) = n!/((r!(n-r)!)

N1 = 6 choose 5 = C(6,5) = 6!/(5!/1!) = 6 ways

(same as number of ways to choose 1 reindeer to be left out).

N2 = 5! ways (5 choices for the first, 4 choices for the second, 3 for the third, and 2 for the fourth, and 1 for the last) = 5*4*3*2*1 = 120 ways.

So total number of arrangements

= N1 * N2 = 6 * 120 = 720 ways.

Alternatively, you can line up the 5 vacant spaces and choose the first among 6 reindeer, second among 5, third among 4, fourth among 3, and the last one among 2 for a total of

6*5*4*3*2 = 720 arrangements.

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Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

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Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

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

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 average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

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Step-by-step explanation:

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