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3 years ago

13

About how many laps around the perimeter of the field would equal 1 mile? Explain

1 answer:

MariettaO [177]3 years ago

8 0

<span>So if you run around the outside of the enfield, it would take a little more than 5 laps to reach one mile. If you cut across at the goal line, you lap would only be 306.6 yards and it would take 5.74 laps to reach 1 mile </span>

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Rewrite each expression as a sum <br> -8x^2 + 3xy - 9x - 3

Gnesinka [82]

$-8x^2 + 3xy - 9x - 3
$

$There \ are \ no \ like \ terms.$

$No \ Solution$

3 0

3 years ago

horsena [70]

Answer:

1) 102.7 meters

2) 11

Step-by-step explanation:

1) -4.9x² + 75x = -4.9x² + 50x + 38

25x = 38

x = 1.52 s

Height = -4.9(1.52)² + 75(1.52) = 102.67904 m = 102.7 m

2) f(x) = x² + 3x - 2

f(2) = 2² + 3(2) - 2 = 8

f(6) = 6² + 3(6) - 2 = 52

Average rate of change:

(52-8)/(6-2) = 11

5 0

3 years ago

Read 2 more answers

If 5-3(2a+1)=-4a+10, what is the value of a+2

AlekseyPX

Answer:

-2

<em>BRAINLIEST, PLEASE!</em>

Step-by-step explanation:

5 - 3(2a + 1) = -4a + 10

5 - 6a - 3 = -4a + 10

2 - 6a = -4a + 10

-2a = 8

a = -4

-4 + 2 = -2

5 0

3 years ago

Read 2 more answers

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

$SE_{p_A-p_B} = \sqrt{0.00274}$

$SE_{p_A-p_B} = 0.052$

Solving (a): Standard error using bootstrapping.

Following the below steps.

Open Statkey
Under Randomization Hypothesis Tests, select Test for Difference in Proportions
Click on Edit data, enter the appropriate data
Click on ok to generate samples
Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A Sample B

$x_A = 23$ $x_B = 57$

$n_A = 100$ $n_B =250$

$p_A = 0.230$ $p_A = 0.228$

So, we have:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}$

$SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}$

$SE_{p_A-p_B} = \sqrt{0.002475064}$

$SE_{p_A-p_B} = 0.050$

5 0

3 years ago

What’s the answer to<br> cos-1(168/240)

Nezavi [6.7K]

Answer:

ans= 45.6

Step-by-step explanation:

I think the ans is 45.6

I hope it will help u...

7 0

3 years ago