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2 years ago

Can y’all please help me ……

Mathematics

1 answer:

Goryan [66]2 years ago

5 0

Answer:

Step-by-step explanation:

( 8 ). m∠1 = m∠ 4 = m∠ 5 = m∠6 = 52°

( 9 ). m∠ 2 = m∠ 3 = m∠ 7 = m∠ 8 = 180° - 52° = 128°

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15. In the diagram AB is parallel to CD. Calculate the value of a. NOT TO SCALE

VikaD [51]

Same-side interior angles, which are shown here, are supplementary, which means that they add up to 180 degrees.

5a + a = 180

6a = 180

a = 30 degrees

Hope this helps! :)

5 0

3 years ago

A child lies on the ground and looks up at the top of a 17-ft tree nearby. The child is 14 ft away from the tree. What is the an

MakcuM [25]

The value of a is 63.43 degrees

6 0

3 years ago

Read 2 more answers

Graph the line y-2=3/10(x-6)

fenix001 [56]

Answer:fbrhthtjthth

Step-by-step explanation:

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3 0

3 years ago

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p: n = 195,

Troyanec [42]

Answer:

c. 0.778 < p < 0.883.

Step-by-step explanation:

The formula for confidence interval for proportion =

p ± z score × √p(1 - p)/n

p = x/n

n = 195, x = 162

z score for 95% confidence Interval = 1.96

p = 162/195

p = 0.8307692308

p ≈ approximately equal to = 0.8308

0.8308 ± 1.96 × √0.8308 × (1 - 0.8308)/195

0.8308 ± 1.96 ×√0.8308 × 0.1692/195

0.8308 ± 1.96 × √0.0007208788

0.8308 ± 1.96 × 0.0268491862

0.8308 ± 0.052624405

Confidence Interval

= 0.8308 - 0.052624405

= 0.778175595

Approximately = 0.778

= 0.8308 + 0.052624405

= 0.883424405

Approximately = p

0.883

Therefore, the confidence interval for this proportion = (0.778, 0.883) or option c. 0.778 < p < 0.883

6 0

3 years ago

Please I Really Need Help.The coordinate plane below represents a city. Points A through F are schools in the city.

almond37 [142]

Answer:

Part A) The system of inequalities is

$x\geq2$ and $y\geq2$

Part B) In the procedure

Part C) The schools that Natalie is allowed to attend are A,B and D

Step-by-step explanation:

Part A: Using the graph above, create a system of inequalities that only contain points C and F in the overlapping shaded regions

we have

Points C(2,2), F(3,4)

The system of inequalities could be

$x\geq2$ -----> inequality A

The solution of the inequality A is the shaded area at the right of the solid line x=2

$y\geq2$ -----> inequality B

The solution of the inequality B is the shaded area above of the solid line y=2

see the attached figure N 1

Part B: Explain how to verify that the points C and F are solutions to the system of inequalities created in Part A

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Verify point C

C(2,2)

Inequality A

$x\geq2$ -----> $2\geq2$ ----> is true

Inequality B

$y\geq2$ ------> $2\geq2$ ----> is true

therefore

Point C is a solution of the system of inequalities

Verify point D

F(3,4)

Inequality A

$x\geq2$ -----> $3\geq2$ ----> is true

Inequality B

$y\geq2$ ------> $4\geq2$ ----> is true

therefore

Point D is a solution of the system of inequalities

Part C: Natalie can only attend a school in her designated zone. Natalie's zone is defined by y < −2x + 2. Explain how you can identify the schools that Natalie is allowed to attend.

we have

$y < -2x+2$

The solution of the inequality is the shaded area below the dotted line $y=-2x+2$

The y-intercept of the dotted line is the point (0,2)

The x-intercept of the dotted line is the point (1,0)

To graph the inequality, plot the intercepts and shade the area below the dotted line

see the attached figure N 2

therefore

The schools that Natalie is allowed to attend are A,B and D

3 0

3 years ago