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3 years ago

6

Which line on the graph represents the equation y + 4 = -2x/3

1 answer:

Korolek [52]3 years ago

3 0

Answer:

The answer is line D

Step-by-step explanation:

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When do you can ride her bike .8 miles in six minutes how many miles can she ride her bike and 1.5 hours

Dafna11 [192]

They can ride their bike 12 miles because in 6 minutes they ride 0.8 miles so in 1.5 hours they ride 12 miles. 0.8 * 10 = 8 miles 1 hour so 0.8 * 5 = 4 miles 0.5 hours 8 miles + 4 miles =12 miles 1 hour + 0.5 hours = 1.5 hours

7 0

3 years ago

Two rectangular picture frames have the same area of 45 square inches but have different side lengths. Frame A has a length of 6

AnnyKZ [126]

Ima say B B B B B
That is what I think.

5 0

3 years ago

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

From a radar station, the angle of elevation of an approaching airplane is 32.5 degree. The horizontal distance between the plan

denis23 [38]

Answer: 42.21 km

Step-by-step explanation:

We can solve this using trigonometry, since we have the following data:

$\theta=32.5\°$ is the the angle of elevation

$d=35.6 km$ is the horizontal distance between the plane and the radar station

$x$ is the hypotenuse of the right triangle formed between the radar station and the airplane

Now, the trigonometric function that will be used is <u>cosine</u>:

$cos\theta=\frac{d}{x}$ because $d$ is the adjacent side of the right triangle

$cos(32.5\°)=\frac{35.6 km}{x}$

Finding $x$ :

$x=\frac{35.6 km}{cos(32.5\°)}$

$x=42.21 km$

4 0

4 years ago

A boat sails on a bearing of 77 degrees for 135 miles and then turns and sails 207 miles on a bearing of 192 degrees. Find the d

Andrew [12]

Answer:

193.53 miles

Step-by-step explanation:

Please see the diagram for understanding of how the angles were derived,

Applying Alternate Angles, ABO =77 degrees

The bearing from B to C is 192=180+12 degrees

Subtracting 12 from 77, we obtain the angle at B as 65 degrees.

We want to determine the boat's distance from its starting point.

In the diagram, this is the line AC.

Applying Law of Cosines:

$b^2=a^2+c^2-2acCosB\\b^2=207^2+135^2-2(207)(135)Cos65^\circ\\b^2=37453.8654\\b=\sqrt{37453.8654} \\b=193.53\: miles$

The distance of the boat from its starting point is 193.53 miles (correct to 2 decimal places).

6 0

3 years ago