All categories

3 years ago

Need help please don’t know what to do here

Mathematics

1 answer:

kotykmax [81]3 years ago

4 0

Answer:

Down below

Step-by-step explanation:

a. The range of y = sinθ is [-1,1]

b. The period of y = cosθ is 2π

c. The asymptotes of y = tanθ are -π2, π2, πn

d. The amplitude of y = sinθ is 1

e. The period of y = tanθ is π

f. The max value of y = cosθ is 1

You might be interested in

A stockbroker has money in three accounts. The interest rates on the three accounts are 7%, 8%, 9%. If she has twice as much mon

Sveta_85 [38]

Answer:

x=amount invested at 7%

2x=amount invested at 8%

3x=amount invested at 9%

interest=principal*rate*time (time=1 year)

$150=0.07x+0.08*2x+0.09*3x

$150=0.07x+0.16x+0.27x

$150=0.50x

$1500=5x

x=$300 invested at 7%

2x=$600 invested at 8%

3x=$900 invested at 9%

Step-by-step explanation:

5 0

3 years ago

F (x) = -4x² + 10x + 13

Alexus [3.1K]

Answer:

f(7) = -113

Step-by-step explanation:

Hello!

Substitute 7 for x in the equation.

<h3>Evaluate</h3>

f(x) = -4x² + 10x + 13
f(7) = -4(7)² + 10(7) + 13
f(7) = -4(49) + 70 + 13
f(7) = -196 + 70 + 13
f(7) = -113

f(7) is -113.

3 0

2 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

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

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

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

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

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

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

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

I need help with this

DochEvi [55]

Answer: B. 8 x 10^3

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

PLSSS HELP IF YOU TURLY KNOW THISS

Doss [256]

Answer:

the answer s 6 and seventh squared so it's a 7 on the top of the 6

Step-by-step explanation:

first what your do is decide 6'5 and 6'2 and you would get 6'3. so just multiply 6'3 and 6'4 and you should get 6'7

7 0

2 years ago