Help me wit this...ion even know what these r

Answers

Answer 1

Answer:

Answer:

Curl up test

Muscular endurance is the ability of a muscle or group of muscles to sustain repeated contractions against a resistance for an extended period of time. It is one of the components of muscular fitness, along with muscular strength and power.

Answer 2

Answer: Hi thank u by hiiiiiii

Related Questions

Odessa starts counting the frogs in the small pond that the forest service just set up by her house. She marks the data in this graph. If y equals the number of frogs and x equals the number of months that have passed, the frog population can be described by the mathematical formula y = x.
A fire burned a square - shaped forest 14mi on a side the area is ?
What is 5\8 times 16
What percent of 68 is 34
What is the solution to the equation 9x+27=9( x+2) +9

Solve for the variable 8x + 15 = 63

Answers

Answer:

Step-by-step explanation:

63-15 =48 48/8 is 6

Answer:

x = 6

Step-by-step explanation:

8x + 15 = 63

8x = 63 - 15

8x = 48

x = 48/8

x = 6

Is line / parallel to line m? Explain.
Note: Not drawn to scale

Answers

Parallel lines are defined to be two lines that will never intersect even when they go on infinitely. Think of it like railroad tracks. The rails must always be parallel, or the train will gets stuck. By observing lines L and M, you'll notice both lines make perfect railroad tracks, and they will never intersect. These lines are parallel.

Alex buys six equally priced candy bars for $9:00 what is the unit rate?

Answers

Answer:

$1.5 unit

Step-by-step explanation:

According to the given situation, the calculation of the unit rate is shown below:-

Candy bars = 6

Amount of candy bars = $9:00

Unit rate

$= (Amount\ of\ candy\ bars)/(Number\ of\ candy\ bars)$

Now we will put the values into the above formula

$= (\$9:00)/(6)$

Which gives result

= $1.5 unit

Therefore for computing the unit rate we simply divide the amount of candy bars by the number of candy bars.

A meteorologist is studying the speed at which thunderstorms travel. A sample of 10 storms are observed. The mean of the sample was 12.2 MPH and the standard deviation of the sample was 2.4. What is the 95% confidence interval for the true mean speed of thunderstorms?

Answers

Answer:

The 95% confidence interval for the true mean speed of thunderstorms is [10.712, 13.688].

Step-by-step explanation:

Given information:

Sample size = 10

Sample mean = 12.2 mph

Standard deviation = 2.4

Confidence interval = 95%

At confidence interval 95% then z-score is 1.96.

The 95% confidence interval for the true mean speed of thunderstorms is

$CI=\overline{x}\pm z*(s)/(√(n))$

Where, $\overline{x}$ is sample mean, z* is z score at 95% confidence interval, s is standard deviation of sample and n is sample size.

$CI=12.2\pm 1.96(2.4)/(√(10))$

$CI=12.2\pm 1.487535$

$CI=12.2\pm 1.488$

$CI=[12.2-1.488, 12.2+1.488]$

$CI=[10.712, 13.688]$

Therefore the 95% confidence interval for the true mean speed of thunderstorms is [10.712, 13.688].

Please help 30 points plus brainlyist who does firstDetermine which expressions can be simplified further, and which cannot. Sort the expressions into the correct
category
2x + 3y
Can Be Simplified
Cannot Be Simplified
x + x
4r +
7y + 1
4y + 4x
y + 2y

Answers

Answer:

can - y +2y

9x+6x

4x+x

can't 4y+4x

7y+1

2x+3y

Answer:

Step-by-step explanation:

expresion can be simplified is they have like terms such as

x+x=2x

y+2y=3y

expresions can NOT be simplified if they have difrerent variables or just one number suchh as

2x+3y

7y+1

4y+4x

I do not know what is 4r+

Sample annual salaries (in thousands of dollars) for employees at a company are listed. 51 53 48 62 34 34 51 53 48 30 62 51 46 (a) Find the sample mean and sample standard deviation. (b) Each employee in the sample is given a $5000 raise. Find the sample mean and sample standard deviation for the revised data set. (c) Each employee in the sample takes a pay cut of $2000 from their original salary. Find the sample mean and the sample standard deviation for the revised data set. (d) What can you conclude from the results of (a), (b), and (c)?

Answers

Answer:

Mean increase or decrease (same quantity) according to the quantity of the increment or reduction

As all elements were equally affected the standard deviation will remain the same

Step-by-step explanation:

For the original set of salaries: ( In thousands of $ )

51, 53, 48, 62, 34, 34, 51, 53, 48, 30, 62, 51, 46

Mean = μ₀ = 47,92

Standard deviation = σ = 9,56

If we raise all salaries in the same amount ( 5 000 $ ), the nw set becomes

56,58,53,67,39,39,56,58,53,35,67,56,51

Mean = μ₀´ = 52,92

Standard deviation = σ´ = 9,56

And if we reduce salaries in the same quantity ( 2000 $ ) the set is

49,51,46,60,32,32,49,51,46,28,60,49,44

Mean μ₀´´ = 45,92

Standard deviation σ´´ = 9,56

What we observe

1.-The uniform increase of salaries, increase the mean in the same amount

2.-The uniform reduction of salaries, reduce the mean in the same quantity

3.-The standard deviation in all the sets remains the same.

We can describe the situation as a translation of the set along x-axis (salaries). If we normalized the three curves we will get a taller curve (in the first case) and a smaller one in the second, but the data spread around the mean will be the same

Any uniform change in the data will directly affect the mean value

Uniform changes in values in data set will keep standard deviation constant

Final answer:

The mean salary is affected by each employee's changes in salary, such as raises and pay cuts, but the standard deviation (the spread of salaries) remains the same provided the change is the same for all individuals.

Explanation:

To answer this question, we need to calculate the sample mean and sample standard deviation in each case. The sample mean is the average of the data, while the sample standard deviation is a measure of the amount of variation or dispersion in the data set.

(a) Calculate the sample mean and sample standard deviation of the initial salaries. This involves summing all the salaries and dividing by the total number to get the mean, then calculating the standard deviation using the formula: square root of [sum of (each salary - mean salary)² divided by (total number of salaries - 1)].
(b) When each employee is given a $5000 raise, the mean will increase by 5, while the standard deviation will remain the same because raises do not affect the dispersion of the team's salaries.
(c) Similarly, when each employee takes a pay cut of $2000, the mean salary will decrease by 2, but the standard deviation will again remain the same because pay cuts do not affect the dispersion of the team's salaries.
(d) We can conclude that while the mean is affected by changes in individual salaries (like raises and pay cuts), the standard deviation is not, provided the change is the same for all individuals. Therefore, it shows that while supply changes can affect the central tendency (mean), they do not impact how spread out the salaries are (standard deviation).

Learn more about Mean and Standard Deviation here:

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What sample size would be required to estimate the true proportion of American female business executives who prefer the title "Ms.," with an error of ±0.025 and 98 percent confidence? (Enter your answer as a whole number (no decimals). Use a z-value taken to three decimal places in your calculations.) Sample size

A vending machine dispenses hot chocolate or coffee. Service time is 20 seconds per cup and is constant. Customers arrive at a mean rate of 64 per hour, and this rate is Poisson-distributed.

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Below are the jersey numbers of 11 players randomly selected from a football team. Find the range, variance, and standard deviation for the given sample data. What do the results tell us? 33 29 97 56 26 78 83 74 65 47 58 Range = (Round to one decimal place as needed.) Sample standard deviations ? = (Round to one decimal place as needed) Sample variance = (Round to one decimal place as needed) What do the results tell us? a. Jersey numbers are nominal data that are just replacements for names, so the resulting statistics are meaningless. b. Jersey numbers on a football team vary much more than expected c. The sample standard deviation is too large in comparison to the range. d. Jersey numbers on a football team do not vary as much as expected.