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3 years ago

Name the quadrant in which sinθ < 0 and cosθ > 0. I II III IV

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Quadrant IV
at quadrant IV, sinA is negative value while cosA is positive value

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Suppose that the hourly wages of fast food workers are normally distributed with an unknown mean and standard deviation. The wag

Nutka1998 [239]

Answer: 1.303639

Step-by-step explanation:

The t-score for a level of confidence $(1-\alpha)$ is given by :_

$t_{(df,\alpha/2)}$ , where df is the degree of freedom and $\alpha$ is the significance level.

Given : Level of significance : $1-\alpha:0.80$

Then , significance level : $\alpha: 1-0.80=0.20$

Sample size : $n=40$

Then , the degree of freedom for t-distribution: $df=n-1=40-1=39$

Using the normal t-distribution table, we have

$t_{(df,\alpha/2)}=t_{39,0.10}=1.303639$

Thus, the t-score should be used to find the 80% confidence interval for the population mean =1.303639

8 0

3 years ago

Solve the following.<br> 2x^2-7x-4/6x^2+7x+2<0<br> Can someone explain how to solve this.

oksano4ka [1.4K]

Answer:

Which one have same variable gather or decrease up

Step-by-step explanation:

2x^2-4.6x^2=1.4x^2

-7x+7x=0

1.4x^2+2<0

x<0

4 0

3 years ago

Whats the equation of the line that passes threw -4,4 and is parrell to the line y=1/2-4

vovangra [49]

Answer:

$y=\frac{1}{2}x+6$

Step-by-step explanation:

Given:

The equation of the known line is:

$y=\frac{1}{2}x-4$

A point on the unknown line is (-4, 4)

Now, since the two lines are parallel, their slopes must be equal.

Now, slope of the known line is the coefficient of 'x' which is $\frac{1}{2}$ .

Therefore, the slope of the unknown line is also $m=\frac{1}{2}$

Now, for a line with slope 'm' and a point on it $(x_1,y_1)$ is given as:

$y-y_1=m(x-x_1)$

Here, $m=\frac{1}{2}, x_1=-4,y_1=4$ . Therefore,

$y-4=\frac{1}{2}(x-(-4))\\\\y-4=\frac{1}{2}(x+4)\\\\y-4=\frac{1}{2}x+2\\\\y=\frac{1}{2}x+2+4\\\\y=\frac{1}{2}x+6$

Hence, the equation of the unknown line is $y=\frac{1}{2}x+6$ .

5 0

3 years ago

Can someone please help me with this problem? You need to find the value of x and y by using the strategy of ELIMINATION. Please

Setler [38]

Answer:

x = 136/35; y = -⁹/₁₀

Step-by-step explanation:

(1) 7x + 8y = 20

(2) 7x – 2y = 29 Subtract (2) from (1)

10y = -9 Divide each side by 10

(3) y = -⁹/₁₀ Substitute (3) into (1)

7x - 2(-⁹/₁₀) = 29

7x + 18/10 = 29 Subtract 18/10 from each side

7x = 29 - 18/10

7x = (290 - 18)/10

7x = 272/10 Divide each side by 7

x = 272/70

x = 136/35

x = 136/35; y = -⁹/₁₀

Check:

(1) 7(136/35) + 8(-⁹/₁₀) = 20

136/5 - 72/10 = 20

136/5 - 36/5 = 20

100/5 = 20

20 = 20

(2) 7(136/35) – 2(-⁹/₁₀) = 29

136/5 + 18/10 = 29

136/5 + ⁹/₅ = 29

145/5 = 29

29 = 29

5 0

3 years ago

Helpppppppppppppppppppppppp

pogonyaev

Answer:

I'm not sure about "11" but 12. is "by the definition of right angles", and 16. is "by the SAS(side angle side) congruency theorem"

3 0

3 years ago