May someone pls help with this?

Help me please like plase

Vinil7 [7]

Answer:

Adjacent angles.

Step-by-step explanation:

a and b are next to each other on a straight line.

6 0

2 years ago

I want to know the answer of all of them

Nady [450]

A. 3/5 * 60 minutes = 180/5 = 36 minutes
b. (3/4)/12 = 3/(4*12) = 3/48 = 1/16 cup
c. (5 1/3) / (2/3) = (16/3) / (2/3) = 16/3 * 3/2 = 48/6 = 8 days
d. 137/16 = 8.5625 pounds

5 0

3 years ago

The temperature at 10 A.M is 12 degrees Fahrenheit. The temperature at 6:00 A.M. was -7 degrees Fahrenheit. How many degrees did

scoundrel [369]

12 degrees Fahrenheit

7 0

3 years ago

Read 2 more answers

A manufacturer of paper coffee cups would like to estimate the proportion of cups that are defective (tears, broken seems, etc.)

Maksim231197 [3]

Answer:

a

The 95% confidence interval is $0.0503 < p < 0.1297$

b

The sample proportion is $\r p = 0.09$

c

The critical value is $Z_{\frac{\alpha }{2} } = 1.96$

d

The standard error is $SE =0.020$

Step-by-step explanation:

From the question we are told that

The sample size is n = 200

The number of defective is k = 18

The null hypothesis is $H_o : p = 0.08$

The alternative hypothesis is $H_a : p > 0.08$

Generally the sample proportion is mathematically evaluated as

$\r p = \frac{18}{200}$

$\r p = 0.09$

Given that the confidence level is 95% then the level of significance is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the standard of error is mathematically represented as

$SE = \sqrt{\frac{\r p (1 - \r p)}{n} }$

substituting values

$SE = \sqrt{\frac{0.09 (1 - 0.09)}{200} }$

$SE =0.020$

The margin of error is

$E = Z_{\frac{ \alpha }{2} } * SE$

=> $E = 1.96 * 0.020$

=> $E = 0.0397$

The 95% confidence interval is mathematically represented as

$\r p - E < \mu < p < \r p + E$

=> $0.09 - 0.0397 < \mu < p < 0.09 + 0.0397$

=> $0.0503 < p < 0.1297$

7 0

3 years ago

HELP WITH THESE QUESTIONS PART 2.

polet [3.4K]

Answer: (a) IV (b) positive (c) $\frac{\pi}{4}$ (d) C (e) $\sqrt{2}$

Step-by-step explanation:

a) $\frac{15\pi}{4}$ - $\frac{8\pi}{4}$ = $\frac{7\pi}{4}$ , which is located in Quadrant IV per the Unit Circle.

b) sec is $\frac{1}{cos}$ . Cos is the x-coordinate. The x-coordinate inQuadrant IV is positive.

c) the reference angle is the angle from $\frac{7\pi}{4}$ to 2π = $\frac{\pi}{4}$

d) since the angle is below the x-axis and the reference angle is $\frac{\pi}{4}$ , then the angle is equal to $-sec(\frac{\pi}{4}$ )

e) sec = $\frac{1}{cos}$ ⇒ sec $\frac{7\pi}{4}$ = $\frac{2}{\sqrt{2} }$ = $\frac{2}{\sqrt{2} }$ $(\frac{\sqrt{2}} {\sqrt{2}})$ = $\frac{2\sqrt{2} }{2}$ = $\sqrt{2}$