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2 years ago

I need help fast please

Mathematics

1 answer:

kotykmax [81]2 years ago

8 0

Answer:

3.4

Step-by-step explanation:

$Interval=\frac{4-2.2}{9}\\Interval=0.2\\therefore :\\ 6 intervals =6(0.2)\\ 6 intervals =1.2\\x=2.2+1.2\\x=3.4$

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mel-nik [20]

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3 years ago

CAN SOMEONE HELP ME FAST?

maksim [4K]

Answer:

221.38

Step-by-step explanation:

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3 years ago

The graph of f(x) = xx is reflected across the x-axis and then across the y-axis to create the graph of function g(x).

V125BC [204]

Now let's examine the statements:

A)The functions have the same range:FALSE the range changed from y ≥ 0 to y ≤ 0

B)The functions have the same domains. FALSE the doman changed from x ≥ 0 to x ≤ 0

C)The only value that is in the domains of both functions is 0. TRUE: the intersection of x ≥ 0 with x ≤ 0 is 0.

D)There are no values that are in the ranges of both functions. FALSE: 0 is in the ranges of both functions.

E)The domain of g(x) is all values greater than or equal to 0. FALSE: it was proved that the domain of g(x) is all values less than or equal to 0.

F)The range of g(x) is all values less than or equal to 0. TRUE: it was proved above.

3 0

4 years ago

Read 2 more answers

To estimate the mean of a population with unknown distribution shape and unknown standard deviation, we take a random sample of

Naddika [18.5K]

Answer:

$t_{critical} = \pm 1.7793$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 64

Sample mean = 22.3

Sample standard deviation = 8.8

We want to estimate 92% confidence interval.

92% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

Degree of freedom = n - 1 = 63

$t_{critical}\text{ at degree of freedom 63 and}~\alpha_{0.01} = \pm 1.7793$

$22.3 \pm 1.7793(\dfrac{8.8}{\sqrt{64}} ) = 22.3 \pm 1.95723 = (20.34,24.26)$

7 0

3 years ago

Use the arc length formula to find the length of the curve y = 4x − 5, −1 ≤ x ≤ 2. check your answer by noting that the curve is

professor190 [17]

Ok, here we go. Pay attention. The formula for the arc length is $AL= \int\limits^b_a { \sqrt{1+( \frac{dy}{dx})^2 } } \, dx$ . That means that to use that formula we have to find the derivative of the function and square it. Our function is y = 4x-5, so y'=4. Our formula now, filled in accordingly, is $AL= \int\limits^2_1 { \sqrt{1+4^2} } \, dx$ (that 1 is supposed to be negative; not sure if it is til I post the final answer). After the simplification we have the integral from -1 to 2 of $\sqrt{17}$ . Integrating that we have $AL= \sqrt{17}x$ from -1 to 2. $2 \sqrt{17}-(-1 \sqrt{17} )$ gives us $3 \sqrt{17}$ . Now we need to do the distance formula with this. But we need 2 coordinates for that. Our bounds are x=-1 and x=2. We will fill those x values in to the function and solve for y. When x = -1, y=4(-1)-5 and y = -9. So the point is (-1, -9). Doing the same with x = 2, y=4(2)-5 and y = 3. So the point is (2, 3). Use those in the distance formula accordingly: $d= \sqrt{(2-(-1))^2+(3-(-9))^2}$ which simplifies to $d= \sqrt{9+144}= \sqrt{153}$ . The square root of 153 can be simplified into the square root of 9*17. Pulling out the perfect square of 9 as a 3 leaves us with $3 \sqrt{17}$ . And there you go!

5 0

3 years ago

I need help fast please​

I need help fast please