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1 year ago

O. Find the domain and range of each of the three restricted functions.

Mathematics

1 answer:

Orlov [11]1 year ago

4 0

The domain and range of the functions sin x, cos x, and tan x will be (-∞, ∞) and [-1, 1], (-∞, ∞) and [-1, 1], and R - (2n + 1)π/2 and (-∞, ∞) respectively.

<h3>What are domain and range?</h3>

The domain means all the possible values of x and the range means all the possible values of y.

The three restricted functions are given below.

f(x) = sin (x)

f(x) = cos (x)

f(x) = tan (x)

Then the domain of the three restricted functions will be

The domain of sin x will be (-∞, ∞).

The domain of cos x will be (-∞, ∞).

The domain of tan x will be R - (2n + 1)π/2.

Then the range of the three restricted functions will be

The range of sin x will be [-1, 1].

The range of cos x will be [-1, 1].

The range of tan x will be (-∞, ∞).

The graph is given below.

More about the domain and range link is given below.

brainly.com/question/12208715

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3 years ago

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3 0

3 years ago

sveta [45]

This question was not written properly

Complete Question

Steve has 54 stamps in his collection of 13 cent, 29 cent, and 45 cent stamps, totaling a value of $15.98. If the number of 13 cent stamps is doubled, the new total value of his stamp collection would be $17.80. Find the number of each type of stamp in his collection.

Answer:

a) The number of 13 cent = 14

b) The number of 29 cent = 24

c) The number of 45 cent = 16

Step-by-step explanation:

We are told he has 54 stamps in total.

These stamps are;13 cent, 29 cent, and 45 cent stamps.

Let's

Number of 13 cent = a

Number of 29 cent = b

Number of 45 cent = c

Hence,

a + b + c = 54........... Equation 1

The total value of the 3 stamps are $15.98.

1 cent = $0.01

We have

0.13a + 0.29b + 0.45c= $15.98 ......... Equation 2

We know from the question that:

13 cent stamps is doubled, the new total value of his stamp collection would be $17.80.

Therefore

2(0.13a) + 0.29b + 0.45c = 17.8

0.26a + 0.29b + 0.45c = 17.8 ......... Equation 3

Combining Equation 2 and 3 together

0.13a + 0.29b + 0.45c= 15.98 ......... Equation 2

0.26a + 0.29b + 0.45c = 17.8 ......... Equation 3

We eliminate b and c by Subtracting Equation 3 from 3

0.13a = 1.82

a = 1.82/0.13

a = 14

a + b + c = 54........... Equation 1

Since a = 14

14 + b + c = 54

b + c = 54 - 14

b + c = 40

c = 40 - b

Substitute 40 - b for c and 14 for a is Equation 2

0.13a + 0.29b + 0.45c= 15.98 ......... Equation 2

0.13(14 ) + 0.29b + 0.45(40 - b) = 15.98

= 1.82 + 0.29b + 18 - 0.45b = 15.98

Collect like terms

1.82 + 18 - 15.98 = 0.45b - 0.29b

3.84 = 0.16b

b = 3.84/0.16

b = 24

Hence: Solving for c

a + b + c = 54........... Equation 1

14 + 24 + c = 54

38 + c = 54

c = 54 - 38

c = 16

Therefore,

The number of 13 cent = 14

The number of 29 cent = 24

The number of 45 cent = 16

8 0

3 years ago

Consider the following hypothesis test.H0:μ1−μ2=0 Ha:μ1−μ2≠0The following results are for two independent samples taken from the

Julli [10]

Answer:

a) $z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

b) $p_v =2*P(z$

c) Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

Step-by-step explanation:

Information given

$\bar X_{1}= 104$ represent the mean for 1

$\bar X_{2}= 106$ represent the mean for 2

$\sigma_{1}= 8.4$ represent the population standard deviation for 1

$\sigma_{2}= 7.6$ represent the population standard deviation for 2

$n_{1}=80$ sample size for the group 1

$n_{2}=70$ sample size for the group 2

z would represent the statistic

Hypothesis to test

We want to check if the two means for this case are equal or not, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

The statistic would be given by:

$z =\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma^2_1^2}{n_1} +\frac{\sigma^2_2^2}{n_2}}}=$ (1)

Part a

Replacing we got:

$z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

Part b

The p value would be given by this probability:

$p_v =2*P(z$

Part c

Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

6 0

3 years ago