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3 years ago

Find the value of each expression using the given information.

Mathematics

1 answer:

Slav-nsk [51]3 years ago

7 0

Recall that $\cos^2\theta+\sin^2\theta=1$ . So

$\sin\theta=\pm\sqrt{1-\cos^2\theta}$

Given that both $\cos\theta$ and $\tan\theta$ , and knowing that $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ , it follows that we should expect $\sin\theta>0$ , so we take the positive root above.

Now

$\sin\theta=\sqrt{1-\left(-\dfrac15\right)^2}=\dfrac{2\sqrt6}5$

Then

$\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{-\frac15}{\frac{2\sqrt6}5}=-\dfrac1{2\sqrt6}$

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What is... 6x-2y=5 3x+2y=-2

artcher [175]

6x-2y=5
x-intercept = 5/6
 y-intercept = 5/-2
3x+2y=-2
 x-intercept = -2/3
y-intercept = -2/2

7 0

3 years ago

Someone please help The table shows values for functions f(x) and g(x) .

viva [34]

The answer to the question is x = 0

6 0

3 years ago

Read 2 more answers

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago

What is the value of the smallest of five consecutive integers if the least minus twice the greates equals -3

Alexeev081 [22]

Answer: (b)

Step-by-step explanation:

Given

There are five consecutive integers and the least minus twice the greatest equals to -3

Suppose $x,x+1,x+2,x+3,x+4$ are the five consecutive integers

According to the question

$\Rightarrow x-2(x+4)=-3\\\Rightarrow x-2x-8=-3\\\Rightarrow -x=8-3\\\Rightarrow x=-5$

option (b) is correct.

4 0

3 years ago

(3.24 Socks in a drawer). In your sock drawer you have 4 blue, 5 gray, and 3 black socks. Half asleep one morning you grab 2 soc

irga5000 [103]

Answer:

a) Probability of ending up wearing 2 blue socks is 1/11.

b) Probability of ending up wearing no grey socks is 7/22.

c) Probability of ending up wearing at least 1 black sock is 5/11.

d) Probability of ending up wearing a green sock is 0.

e) Probability of ending up wearing matching socks is 19/66.

Step-by-step explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

In your sock drawer, you have 4 blue socks, 5 gray socks, and 3 black ones. Half asleep one morning, you grab 2 socks at random and put them on. Find the probability you end up wearing: a) 2 blue socks. b) no gray socks. c) at least 1 black sock. d) a green sock. e) matching socks.

The explanation of the answer is now given as follows:

The following are given in the question:

n(B) = number of Blue socks = 4

n(G) = number of Gray socks = 5

n(K) = number of black socks = 3

Therefore, we have:

n(T) = Total number of socks = n(B) + n(G) + n(K) = 4 + 5 + 3 = 12

To calculate a probability, the following formula for calculating probability is used:

Probability = Number of favorable outcomes / Number of total possible outcomes ……. (1)

Since this is a without replacement probability, we can now proceed as follows:

a) 2 blue socks

P(B) = Probability of ending up wearing 2 blue socks = ?

Probability of first pick = n(B) / n(T) = 4 / 12 = 1 / 3

Since it is without replacement, we have:

Probability of second pick = (n(B) – 1) / (n(T) – 1) = (4 – 1) / (12 – 1) = 3 / 11

P(B) = Probability of first pick * Probability of second pick = (1 / 3) * (3 / 11) = 1 / 11

b) no gray socks.

Number of favorable outcomes = n(B) + n(K) = 4 + 3 = 7

P(No G) = Probability of ending up wearing no gray socks = ?

Probability of first pick = Number of favorable outcomes / n(T) = 7 / 12

Since it is without replacement, we have:

Probability of second pick = (Number of favorable outcomes – 1) / (n(T) – 1) = (7 – 1) / (12 – 1) = 6 / 11

P(No G) = Probability of first pick * Probability of second pick = (7 / 12) * (6 / 11) = 7 / 22

c) at least 1 black sock.

Probability of at least one black sock = 1 - P(No K)

Number of favorable outcomes = n(B) + n(G) = 4 + 5 = 9

Probability of first pick = Number of favorable outcomes / n(T) = 9 / 12 = 3 /4

Since it is without replacement, we have:

Probability of second pick = (Number of favorable outcomes – 1) / (n(T) – 1) = (9 – 1) / (12 – 1) = 8 / 11

P(No K) = Probability of first pick * Probability of second pick = (3 / 4) * (8 / 11) = 24 / 44 = 6 / 11

Probability of at least one black sock = 1 - (6 / 11) = 5 / 11

d) a green sock.

n(Green) = number of Green socks = 0

Since, n(Green) = 0, it therefore implies that the probability of ending up wearing a green sock is 0.

e) matching socks.

This can be calculated using the following 4 steps:

Step 1: Calculation of the probability of matching blue socks

P(matching blue socks) = P(B) = 1 / 11

Step 2: Calculation of the probability of matching gray socks

P(matching green socks) = Probability of matching gray socks = ?

Probability of first pick = n(G) / n(T) = 5 / 12

Since it is without replacement, we have:

Probability of second pick = (n(G) – 1) / (n(T) – 1) = (5 – 1) / (12 – 1) = 4 / 11

P(matching gray socks = Probability of first pick * Probability of second pick = (5 / 12) * (4 / 11) = 20 / 132 = 5 / 33

Step 3: Calculation of the probability of matching black socks

P(matching black socks) = Probability of matching green socks = ?

Probability of first pick = n(K) / n(T) = 3 / 12 = 1 / 4

Since it is without replacement, we have:

Probability of second pick = (n(K) – 1) / (n(T) – 1) = (3 – 1) / (12 – 1) = 2 / 11

P(matching black socks) = Probability of first pick * Probability of second pick = (1 / 4) * (2 / 11) = 2 / 44 = 1 / 22

Step 4: Calculation of the probability of ending up wearing matching socks

P(matching socks) = Probability of ending up wearing matching socks = ?

P(matching socks) = P(matching blue socks) + P(matching grey socks) + P(matching black socks) = 1/11 + 5/33 + 1/22 = (6 + 10 + 3) / 66 = 19/66

6 0

3 years ago