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3 years ago

Which three expressions are equivalent to 10x +11

Mathematics

1 answer:

BARSIC [14]3 years ago

8 0

Answer:

B, C, and E

Step-by-step explanation:

Simplify all answer choices:

5(2x+10)+1=

10x+50+1=

10x+51

7(x+2)+3x-3=

7x+14+3x-3=

10x+11

3(3x+4)+x-1=

9x+12+x-1=

10x+11

2(6x+4)+2x+5=

12x+8+2x+5=

14x+13

2(6x+5)-2x+1=

12x+10-2x+1=

10x+11

Answer choices B, C, and E simplify to 10x+11

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Answer:

k=7

Step-by-step explanation:

first, plug in the given values

2x+3y=k --> 2(2)+3(1)=k

-->4+3=k

-->7=k

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3 years ago

When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p w

deff fn [24]

Answer:

A 90% confidence interval for p will be narrower than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - α) % confidence interval for a population proportion is:

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

Here $\hat p$ is the sample proportion.

The margin of error of the confidence interval is:

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

The MOE is dependent on:

Confidence level
Standard deviation
Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

The critical value of z for 90% confidence level is:

$z_{\alpha/2}= z_{0.10/2}=z_{0.05}=1.645$

And the critical value of z for 99% confidence level is:

$z_{\alpha/2}= z_{0.05/2}=z_{0.05}=1.96$

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for p will be narrower than the 99% confidence interval.

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3 years ago

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3 years ago

Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)

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This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ], slope is 86.3

111) for [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope = ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11) [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111) [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

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