All categories

2 years ago

Y=2x + ? ITS DUE IN 20 MINUTES

Mathematics

2 answers:

TiliK225 [7]2 years ago

8 0

The answer is Y=2x+2

Flauer [41]2 years ago

7 0

Y=2x+2 is the answer

You might be interested in

Which graph shows a negative rate of change for the interval 0 to 2 on the x-axis

Lena [83]

Answer:

-2

Step-by-step explanation:

3 0

3 years ago

A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If

Alinara [238K]

Answer:

Test statistic z= 1.708

Step-by-step explanation:

given that a television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If this is to happen, it will have to move one of its most viewed shows to another slot.

Sample size n = 827

Sample proportion p = $\frac{438}{827} \\=0.5297$

$H_0: p =0.5\\H_a: p >0.5$

(right tailed test)

This is because they would be interested if more than half viewers are ready for the change

Test statistic = Z = p difference/std error

p difference = 0.0297

Std error = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.5*0.5}{827} } \\=0.0174$

Z =1.708

p valu e= 0.9562

Since p >0.05 we fail to reject H0

Test statistic z= 1.708

8 0

3 years ago

Use the given information to find (a) sin(s+t), (b) tan(s+t), and (c) the quadrant of s+t. cos s = - 12/13 and sin t = 4/5, s an

Anton [14]

Answer:

Part a) $sin(s + t) =-\frac{63}{65}$

Part b) $tan(s + t) = -\frac{63}{16}$

Part c) (s+t) lie on Quadrant IV

Step-by-step explanation:

[Part a) Find sin(s+t)

we know that

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

step 1

Find sin(s)

$sin^{2}(s)+cos^{2}(s)=1$

we have

$cos(s)=-\frac{12}{13}$

substitute

$sin^{2}(s)+(-\frac{12}{13})^{2}=1$

$sin^{2}(s)+(\frac{144}{169})=1$

$sin^{2}(s)=1-(\frac{144}{169})$

$sin^{2}(s)=(\frac{25}{169})$

$sin(s)=\frac{5}{13}$ ---> is positive because s lie on II Quadrant

step 2

Find cos(t)

$sin^{2}(t)+cos^{2}(t)=1$

we have

$sin(t)=\frac{4}{5}$

substitute

$(\frac{4}{5})^{2}+cos^{2}(t)=1$

$(\frac{16}{25})+cos^{2}(t)=1$

$cos^{2}(t)=1-(\frac{16}{25})$

$cos^{2}(t)=\frac{9}{25}$

$cos(t)=-\frac{3}{5}$ is negative because t lie on II Quadrant

step 3

Find sin(s+t)

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute the values

$sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})$

$sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})$

$sin(s + t) =-\frac{63}{65}$

Part b) Find tan(s+t)

we know that

tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

step 1

Find tan(s)

$tan(s)=sin(s)/cos(s)$

substitute

$tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}$

step 2

Find tan(t)

$tan(t)=sin(t)/cos(t)$

substitute

$tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}$

step 3

Find tan(s+t)

$tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))$

substitute the values

$tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))$

$tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})$

$tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})$

$tan(s + t) = -\frac{63}{16}$

Part c) Quadrant of s+t

we know that

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

Find the value of cos(s+t)

$cos(s+t) = cos(s) cos(t) -sin (s) sin(t)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute

$cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})$

$cos(s+t) = (\frac{36}{65})-(\frac{20}{65})$

$cos(s+t) =\frac{16}{65}$

we have that

$cos(s+t)=positive$ -----> (s+t) could be in I or IV quadrant

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

therefore

(s+t) lie on Quadrant IV

4 0

3 years ago

After 100 spins, Lincoln calculated the probability that he would spin a prime number to be 0.35. Using the

quester [9]

Given:

Total number of spins = 100

Probability that he would spin a prime number to be 0.35.

To find:

The theoretical probability of spinning a prime number.

Solution:

Total numbers from 1 to 100 = 100

Total prime numbers from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

So, the prime numbers from 1 to 100 = 25

Now, the probability of spinning a prime number is

$\text{probability}=\dfrac{\text{Total prime numbers from 1 to 100}}{\text{Total numbers from 1 to 100}}$

$\text{probability}=\dfrac{25}{100}$

$\text{probability}=0.25$

Therefore, the theoretical probability of spinning a prime number is 0.25.

3 0

3 years ago

SOMEONE HELP PLZ!!!!!!

barxatty [35]

Answer:b

Step-by-step explanation:

7 0

3 years ago