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3 years ago

In this figure, sin(a) = 23

Mathematics

1 answer:

WITCHER [35]3 years ago

4 0

$a$ and $b$ are complementary, or $b=\dfrac\pi2-a$ . So

$\cos b=\cos\left(\dfrac\pi2-a\right)=\sin a$

and so $\cos b=\dfrac23$ .

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(a) Estimate the area under the graph of f(x) = 2 cos(x) from x = 0 to x = π/2 using four approximating rectangles and right end

nirvana33 [79]

first, we are going to define variables as the following:

a = 0
a = π/2
n = 4 rectangles

Δx = [ b - a ] / n ------>Δx = [ π/2 - 0 ] / 4 = π/8

right endpoints :
sum( seq( 4 cos(x) * π/8 , x , 0+π/8 , π/2 , π/8 ) ) = 3.163065 underestimate

left endpoints:
sum( seq( 4 cos(x) * π/8 , x , 0 , π/2-(π/8) , π/8 ) ) = 4.733861 overestimate

the reason because the actual estimate by integral as the following:
π/2
∫ 4cos(x) dx = 4
0

8 0

3 years ago

On reading at An artic research station showed that the temperature was -35 C. What is this temperature in degrees Fahrenheit?

Travka [436]

the answer is -22 F

5 0

3 years ago

A cricket team has a 0.25 chance of losing

Westkost [7]

Answer:

a) 0.125

b) 0.015635

c) 0.00000095367431640625‬

Step-by-step explanation:

a) $0.25^2 = 0.125$

b) $0.25^3 = 0.015625$

c) $0.25^{10} = 0.00000095367431640625\\\\$

8 0

3 years ago

Read 2 more answers

PLEASE HELP WORTH 99 POINTS AND CORRECT ANSWER WILL GET BRAINLIEST

Soloha48 [4]

Hello!

I believe the answers are:

A) The vertex is (1, −6)

C) The graph has a minimum.

I hope it helps!

7 0

3 years ago

Read 2 more answers

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = population proportion.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, the test statistics = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0

3 years ago