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2 years ago

15

Select the correct answer from each drop-down menu.

1 answer:

lesantik [10]2 years ago

7 0

Answer: The sample mean is 14.98

The second one is NOT standard deviation

Step-by-step explanation: Sorry thats all i got but good luck!

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Write an equation for the following real life scenario.

mojhsa [17]

Answer:

1/2x = 16

x = 32

Step-by-step explanation:

5 0

2 years ago

The cost of t-shirts varies directly with the number of t-shirts. The data is shown in the graph. If x = is number of t-shirts,

Lera25 [3.4K]

The answer is y = 5.5x, so B

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3 years ago

Read 2 more answers

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

In a 20% off sale a coat was £220. work out the original price

BARSIC [14]

10% of 220= £22
10% *2 = £44
£220- £44 =£176

5 0

3 years ago

A UFO is coming in to pick up some test subjects. Their elevation above the ground is 60,000 feet. Their laser sights tell them

charle [14.2K]

Answer:

41.83°

Step-by-step explanation:

It is given that :

The vertical distance of the UFO from the ground = 60,000 feet

The diagonal distance of the landing area to the UFO = 90,000 feet

In order to find the angle of depression, we have to use the trigonometric ratios, i.e.,

$\sin \ \theta = \frac{\text{perpendicular distance}}{\text{hypotenus}}$

$\sin \ \theta = \frac{60,000}{90,000}$

$\sin \ \theta = \frac{2}{3}$

$\sin \ \theta = 0.667$

∴ $\theta = \sin^{-1 }(0.667)$

= 41.83°

3 0

3 years ago