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3 years ago

14

Me and my friend were checking answers, but then our answers differed on one question by one digit. The question is: [(21-13)+(3

2-24)]×4

2 answers:

Yanka [14]3 years ago

8 0

[(21-13)+(32-24)]x4

[8+8]4

16x4

64

That’s what I got

Sunny_sXe [5.5K]3 years ago

5 0

[(21-13)+(32-24)] x 4

[(8)+(8)] x 4

16 x 4

= 64

If this helped you, please list me as brainliest!

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Please help me!! thank you!!

I am Lyosha [343]

Answer:

A. 106 + 2x + 50 = 180

Step-by-step explanation:

A. 106 + 2x + 50 = 180. Correct.

B. 2x + 50 = 180. Incorrect. Angle is not a Straight Angle.

C. 2x + 50 = 106. Incorrect. That equals the first angle.

hope this helps.

4 0

2 years ago

A small group of students are aged 12, 14, 12, 15, 14, 13, 12, 12. What is their mean age?

Ilia_Sergeevich [38]

Answer:

12

Step-by-step explanation:

12 is the most number in the equation

5 0

3 years ago

Read 2 more answers

How many times larger is 3×108 than 3×102?

Vinvika [58]

Answer: 120

Step-by-step explanation:

3 x 108 = 324

3 x 102 = 204

so 324 - 204 = 120

4 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

8 0

3 years ago

7 and 1/5 -2 and 3/5 equal

Furkat [3]

7 1/5 - 2 3/5 =
36/5 - 13/5 =
23/5 or 4 3/5 <==

6 0

2 years ago