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2 years ago

Please help with both questions

Mathematics

2 answers:

MakcuM [25]2 years ago

5 0

Refer to the attachment for your answer, just remember the fact that as x is an integer in 2nd question and x > 3.66... so x's smallest value will just be 4, and in 1st question you can use the number line for a better understanding

tresset_1 [31]2 years ago

4 0

Value of n is greater than (-2) and less than or equal to 3.
n = {-1,0,1,2 3}

3x +5 > 16
Subtract 5 from both sides
3x > 16 - 5
3x > 11
x > 11/3
x> 3.67

X= 4 is the smallest value

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I need help with 11 and 12

Yakvenalex [24]

Answer:

Both are inverse pairs

Step-by-step explanation:

Question 11

$g(x)= 4 + \dfrac{8}{5}x$

(a) Rename g(x) as y

$y = 4 + \dfrac{8}{5}x$

(b) Solve for x :

$\dfrac{8}{5}x = y - 4$

(c) Multiply each side by ⅝

$x = \dfrac{5}{8}(y - 4) = \dfrac{5}{8}y - \dfrac{5}{2}$

(d) Switch x and y

$y = \dfrac{5}{8}x - \dfrac{5}{2}$

(e) Rename y as the inverse function

$g^{-1}(x) = \dfrac{5}{8}x - \dfrac{5}{2}$

(f) Compare with your function

$f(x) = \dfrac{5}{8}x - \dfrac{5}{2}\\\\f(x) = g^{-1}(x)$

f(x) and g(x) are inverse functions.

The graphs of inverse functions are reflections of each other across the line y = x.

In the first diagram, the graph of ƒ(x) (blue) is the reflection of g(x) (red) about the line y = x (black)

Question 12

h(x)= x - 2

(a) Rename h(x) as y

y = x - 2

(b) Solve for x:

x = y + 2

(c) Switch x and y

y = x + 2

(e) Rename y as the inverse function

h⁻¹(x) = x + 2

(f) Compare with your function

f(x) = x + 2

f(x) = h⁻¹(x)

h(x) and ƒ(x) are inverse functions.

The graph of h(x) (blue) reflects ƒ(x) (red) across the line y = x (black).

5 0

3 years ago

(NEED HELP WITH NUMBER 1!!!)

alexandr1967 [171]

Answer:

35 ia the answer sorry if im wrong

7 0

3 years ago

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

3 years ago

please help I am so stuck on this! 100 points and brainiest the choices for each question are in the second attachment

77julia77 [94]

Answer:

Step-by-step explanation:

(A) Definition of Supplementary angles

(B) Distributive property

(D) Reflexcive property

(E) Division property of equality

7 0

3 years ago

Read 2 more answers

How does the graph of g(x)= 1/x-5+2 compared to the graph of the parent function f(x)=1/x?

Finger [1]

Answer: The graph is shifted to the right by 5 units and shifted up from the parent function.

Step-by-step explanation:

1. By definition, a parent function is the simplest form of a function.

2. You have the following parent function given in the problem:

$f(x)=\frac{1}{x}$

2. The function g(x) is obtained from the function f(x)

$g(x)=\frac{1}{x-5}+2$

Where the subtraction of 5 indicates the horizotanl shift to the left and the addition of 2 indicates the vertical shift up.

3. Therefore, keeping the information above on mind, you can conclude that the answer is: the graph is shifted to the right by 5 units and shifted up from the parent function.

3 0

4 years ago

Please help with both questions ​

Please help with both questions