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2 years ago

Please show me how to do this problem

Mathematics

1 answer:

sleet_krkn [62]2 years ago

5 0

Answer:

Hi there, you can use Pitagoras relationship to solve this up (Without using any other "tool".

Point A is located wheter in (-6,8) or could be located in (-6,8)

Step-by-step explanation:

Point A is 10 units away from the origin, and its x coordinate is -6, using this information, you can draw a triangle from 0 to -6 (the basis) and using 10 as the hypotenuse, so, you have this equation:

$H^{2} = x^{2} +y^{2}$

solving for Y variable, you have this

y=± $\sqrt{10^{2}-(-6)^{2} }$

Note the combined sing ±, meaning that y=8 is a valid answer and that y= -8 is also a valid answer.

Hope It helps

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A machine can stamp 75 caps per minute. At this rate how long will it take to stamp 3000?

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<h2>Answer:</h2><h2>40 minutes</h2>

Step-by-step explanation:

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3 years ago

Read 2 more answers

A store has 263 board games. Sit has 100 fewer puzzles than board games. The store has 10 more action figures than puzzles. Writ

Yuri [45]

so is it has 100 puzzles less than board games and you have 263 board games just subtract 263-100 which is 163. if it has 10 more action figures than puzzles and you have 163 puzzles you would add 10 to 163 which would give you 173 action figures.

163 puzzles and 173 action figures is the answer

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3 years ago

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Evaluate f(-3) for f(x) = 2^x a. 1/8 b. -8 c. 8 d. -6

pickupchik [31]

Answer:

Step-by-step explanation:

f(-3) = 2^-3

to get rid of a negative in an exponent, move the whole thing to the denominator.

1/(2^3)

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1/8

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1 year ago

Error Analysis your friend evaluate (10-5)/5 and gets 9 for an answer. Explain your friend's error

alexandr402 [8]

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3 years ago

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

yKpoI14uk [10]

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.

3 0

2 years ago