Now just 2. Thank you

Grade 4c left their school at7:30 am for their trip to the market. They return at 10:45 am how much time did they spend onthe tr

julia-pushkina [17]

Answer:

3hrs 15 minutes

Step-by-step explanation:

10:45

-07:30

------------

03:15

------------

3 0

2 years ago

I need help with this proof.<br>

mylen [45]

Answer:

My proof is in the explanation.

Step-by-step explanation:

This is a two-column proof.

One column for statements and the other for the reason for that statement.

Hopefully it shows up well on your screen. Let me know if it doesn't.

Statement | Reason

1) CD is the perpendicular 1) Given

bisector of AB

2) AD=DB 2) Definition of bisector

3) CD=CD 3) Reflexive property

4) mAngleCDA=90 4) Definition of perpendicular

5) mAngleCDB=90 5) Definition of perpendicular

6) mAngleCDA=mAngleCDB 6) Substitution property

7) Corresponding parts of each 7) SAS

triangle are congruent (side-angle-side)

8) AC=CB 8) The two triangles are ............................................................................congruent so

the corresponding parts ............................................................................are

congruent.

4 0

2 years ago

The mean and standard deviation of a random sample of n measurements are equal to and , respectively. a. Find a % confidence i

marusya05 [52]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$33.55 < \mu < 35.5$

b

$34.03 < \mu < 34.969$

c

Generally the width at n = 49 is mathematically represented as

$w = 2 * E$

$w = 2 * 0.952$

$w = 1.904$

Generally the width at n = 196 is mathematically represented as

$w = 2 * E$

$w = 2 * 0.4687$

$w = 0.9374$

d

The correct option is E

Step-by-step explanation:

From the question we are told that

The sample mean is $\= x = 34.5$

The standard deviation is $s = 3.4$

Generally given that the confidence level is 95% then the level of significance is

$\alpha = (100 - 95)\%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Considering question a

From the question n = 49

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }$

=> $E = 1.96* \frac{ 3.4 }{\sqrt{49} }$

=> $E = 0.952$

Generally 95% confidence interval is mathematically represented as

$\= x -E < p < \=x +E$

$34.5 -0.952 < p < 34.5 + 0.952$

=> $33.55 < \mu < 35.5$

Considering question b

From the question n = 196

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }$

=> $E = 1.96* \frac{ 3.4 }{\sqrt{196} }$

=> $E = 0.4687$

Generally 95% confidence interval is mathematically represented as

$\= x -E < p < \=x +E$

$34.5 -0.4687 < p < 34.5 +0.4687$

=> $34.03 < \mu < 34.969$

Considering question c

Generally the width at n = 49 is mathematically represented as

$w = 2 * E$

$w = 2 * 0.952$

$w = 1.904$

Generally the width at n = 196 is mathematically represented as

$w = 2 * E$

$w = 2 * 0.4687$

$w = 0.9374$

Now when the sample size is quadrupled i.e from n = 49 to n = 196

The width of the confidence interval decrease by 2 from 1.904 to 0.9374

7 0

3 years ago

Help please? I’m not so intelligent.

nlexa [21]

Answer:

It is 33. Answer B is correct.

5 0

3 years ago

Brainliest if correct

slava [35]

Answer:

-8

2

12

52

Step-by-step explanation:

y = 10x + 2

x = -1, y = 10(-1) + 2 = -8

x - 0, y = 0 + 2 = 2

x = 1, y = 10 + 2 = 12

x = 5, y = 10*5 + 2 = 52

5 0

2 years ago