All categories

3 years ago

How do I solve these equations?

Mathematics

1 answer:

zimovet [89]3 years ago

3 0

Can you use a calculator? If yes, then isolate 7) cot 8) tan 2 theta, etc. Then, input the inverse of that trig into the calculator.

Ex. #7 cot theta= -6/5
Then, input arccot (-6/5) (aka inverse cot (-6/5)) into your calculator.

You might be interested in

Please answer <br>Find the product.<br>(7m^2-3m-7)(m^2-7m-6)

densk [106]

Step-by-step explanation:

(7m²-3m-7)(m²-7m-6) =

7m⁴-49m³-42m²-3m³+21m²+18m

-7m²+49m+42

= 7m⁴-52m³-28m²+67m+42

6 0

3 years ago

The length of a rectangle is one foot more than twice its width. If the area of the rectangle is 300ft2, find the dimensions of

Gemiola [76]

Answer:

12 ft x 25 ft

Step-by-step explanation:

If L is the length and W is the width:

L = 2W + 1

300 = LW

Substitute:

300 = (2W + 1) W

300 = 2W² + W

0 = 2W² + W − 300

0 = (2W + 25) (W − 12)

W = 12

L = 2W + 1

L = 25

The rectangle is 12 feet wide and 25 feet long.

5 0

3 years ago

18. A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be

avanturin [10]

Answer:

a. =50 ± 4.67

b. =50 ± 4.81

decreasing the sample size increase the margin of error

c. =50 ± 3.51

decreasing the confidence level reduced the margin of error

d. No. because the confidence interval coefficient is gotten from the table of normal distribution of Z value. therefore the confidence interval is only used for normal distributed population

Step-by-step explanation:

given

The sample mean, x, = 50,

the sample standard deviation, s, = 8.

a. Construct a 98% confidence interval for m if the sample size, n, is 20

for a 98% confidence interval of an inﬁnite population, thus for a sample of size N and mean x, drawn from an inﬁnite population having a standard deviation of σ, the mean value of the population is estimated to by:

x± 2.33σ /√N

for a conﬁdence level of 98%, Zc = 2.33 gotten from the table of confidence interval.

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√20) and 50+ ( 2.33*8 /√20)

with 98% conﬁdence in this prediction.

=50 ± 4.67

=45.33 or 54.67

(b) Construct a 98% confidence interval for m if the sample size, n, is 15.

using the same method as above

x± 2.33σ /√N

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√15) and 50+ ( 2.33*8 /√15)

=50 ± 4.81

=45.19 or 54.81

decreasing the sample size increase the margin of error

(c) Construct a 95% confidence interval for m if the sample size, n, is 20. Compare the results to those obtained in part

95% confidence interval =1.96 (gotten from table of confidence coefficient)

using x±1.96 σ /√N

This indicates that the mean value of the population lies between

50- (1.96 *8 /√20) and 50+ (1.96 *8 /√20)

=50 ± 3.51

=46.49 or 53.51

c. decreasing the confidence level reduced the margin of error

(d) Could we have computed the confidence intervals in parts (a)–(c) if the population had not been normally distributed?

No. because the confidence interval coefficient is gotten from the table of normal distribution of Z value. therefore the confidence interval is only used for normal distributed population

7 0

3 years ago

What is the area of this triangle in the coordinate plane?

Klio2033 [76]

Hey. Let me help you on this one.

In order for us to solve this problem, we will be needing to find both length and width of the triangle on this coordinate plane. We will also need to know how to use the triangle area formula.

I will be attaching an image containing the shown difference between a rectangle, and a triangle, and on how to understand why the area of a triangle is like this.

First of all, let's find the length of the triangle. It's 11 units, and we can determine that by counting up all of the units, or simply finding the difference between x-point 13 and 2, which is going to be 11.

Same applies to the width of the triangle. We will need to either count up all the units individually, or find a difference between a higher y-point and lower y-point. That difference is going to 5 units, since $9-4=5$ .

Now, let's use the formula for the triangle area and substitute the values.

$S= \frac{a*b}{2}$
$S= \frac{11*5}{2}$
$S= \frac{55}{2}$
$S=27.5$

Answer: Area of the triangle on the coordinate plane is 27.5 units squared.

3 0

4 years ago

Read 2 more answers

While organizing a film festival, you must decide which of the 12 movies will be shown on the big screen. You only have the budg

stealth61 [152]

The equation to calculate how many ways that "r" things can be chosen from a larger set of "n" things is:n! / [r! * (n-r)!] =
12! / (5! * 7!) =
12 * 11 * 10 * 9 * 8 * 7! / (5! * 7!) =12 * 11 * 10 * 9 * 8 / 5 * 4 * 3 * 2 =11 * 9 * 8 =792 ways

http://www.1728.org/combinat.htm

7 0

3 years ago

Read 2 more answers