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4 years ago

Where is the function decreasing? A) 1 < x < ∞ B) 3 < x < ∞ C) -∞ < x < 1 D) -1 < x < ∞

Mathematics

2 answers:

photoshop1234 [79]4 years ago

7 0

Answer:b

Step-by-step explanation:

vlabodo [156]4 years ago

5 0

Answer:

Step-by-step explanation:

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In ΔGHI, the measure of ∠I=90°, the measure of ∠G=62°, and GH = 96 feet. Find the length of IG to the nearest tenth of a foot.

alexandr402 [8]

Answer:

45.1feet

Step-by-step explanation:

Given the following

∠I=90°

∠G=62°, and

GH = 96 feet = Hypotenuse

Required

IG = Adjacent side

Using the SOH CAH TOA identity

Cos theta = Adj/hyp

Cos 62 =IG/96

IG = 96cos62

IG = 96(0.4695)

IG = 45.1feet

Hence the length of IG to the nearest tenth is 45.1feet

6 0

3 years ago

Read 2 more answers

4)In order to set rates, an insurance company is trying to estimate the number of sick daysthat full time workers at an auto rep

bagirrra123 [75]

Answer:

A sample of 18 is required.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.92}{2} = 0.04$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.04 = 0.96$ , so Z = 1.88.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A previous study indicated that the standard deviation was 2.2 days.

This means that $\sigma = 2.2$

How large a sample must be selected if the company wants to be 92% confident that the true mean differs from the sample mean by no more than 1 day?

This is n for which M = 1. So

$M = z\frac{\sigma}{\sqrt{n}}$

$1 = 1.88\frac{2.2}{\sqrt{n}}$

$\sqrt{n} = 1.88*2.2$

$(\sqrt{n})^2 = (1.88*2.2)^2$

$n = 17.1$

Rounding up:

A sample of 18 is required.

3 0

3 years ago

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$

4 0

3 years ago

When using a ruler, where is the unit?

Afina-wow [57]

Answer:

Is it The numbers on the ruler?

Step-by-step explanation:

6 0

2 years ago

F(x) = 4x + 2; find f(8)

likoan [24]

Answer:

f(8)=34

Step-by-step explanation:

To solve, you just substitute x for 8

4(8) + 2

32 + 2

F(8) = 34

8 0

3 years ago