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2 years ago

A,D And E Thats The Answers

Mathematics

2 answers:

Karolina [17]2 years ago

8 0

Answer:

wait what? hmmm hmmmmmm

natali 33 [55]2 years ago

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8. What is the distance around the clock?

Elan Coil [88]

Answer:

The clock face is divided into sixty equal parts, each minute. The minute hand is located on the 20 minute mark at 6:20, the hour hand located between the 30 minute mark and the 35 minute mark. When the minute hand goes all sixty minutes, the hour hand only moves five, so to figure out the location of the hour hand, we look at how much the hour has progressed, in this case 20 minutes, or one third of the hour. So the minute hand has moved one third of the way through the hour, so has the hour hand moved one third of the way through the five minutes, or, five thirds of a minute, which is one and two thirds minute, one minute forty seconds. That puts the hour hand at thirty minutes plus one minute and forty seconds—at 31min 40sec—which is 11min 40sec farther than the minute hand.

Step-by-step explanation:

4 0

2 years ago

What are the zeros of the function f(x) =x^2-11x+24 ?

UNO [17]

Answer:

there is none

Step-by-step explanation:

3 0

3 years ago

7. What polynomial remains after the greatest common factor is factored out of the

lara31 [8.8K]

Answer:the answer is A

Step-by-step explanation:

7 0

2 years ago

Consider the inverse function. Which conclusions can be drawn about f(x) = x2 + 2? Select three options. f(x) has a limited rang

PolarNik [594]

Answer:

f(x) has a limited range

f(x) has a maximum at the point (0, 2)

f(x) has a y-intercept at the point (0, 2).

Step-by-step explanation:

Given the function;

f(x) = x^2+2

The domain is the value of the input variables for which the function will exist. According to the expression given, the function exists on all real values of x. The same goes with range which deals with the output values. It also exists on all real values from 2 and above.

Hence f(x) have a limited range (since values less than 2 are not included compare to domain that exists on all real values) and does not have a restricted domain.

For the x intercept, x intercept occur at y = 0

substitute y = 0 into the function and get y

if y = f(x)

y = x^2+2

0 = x^2 + 2

x^2 = -2

x = 2i

Hence f(x) does not have an x-intercept of (2, 0)

For the y intercept, y intercept occur at x = 0

substitute x = 0 into the function and get y

if y = f(x)

y = x^2+2

y = 0^2 + 2

y = 2

Hence f(x) has a y-intercept at point (0, 2)

f(x) is at maximum if d(fx))/dx = 0

d(fx))/dx = 2x

since d(fx))/dx = 0

0 = 2x

x = 0

substitute x = 0 into the function

f(x) = x^2 + 2

y = 0^2+2

y = 2

Hence f(x) has a maximum at the point (0, 2)

5 0

2 years ago

Read 2 more answers

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

3 years ago