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MAVERICK [17]
3 years ago
7

A vehicle is traveling 540 meters in 6 seconds.How far can it travel in 4 seconds

Mathematics
2 answers:
agasfer [191]3 years ago
7 0
6 seconds = 540 m

-----------------------------
Find 1 second: 
-----------------------------
6 sec = 540 m
1 sec = 540 ÷ 6
1 sec = 90m

-----------------------------
Find 4 seconds:
-----------------------------
1 sec = 90m
4 sec = 90 x 4
4 sec = 360 m

----------------------------------------------------------
Answer: It can travel 360m in 4 seconds.
----------------------------------------------------------
Olenka [21]3 years ago
5 0

Answer:

360m

Step-by-step explanation:

Vehichle=A=540m

time in 540m=60s

1s=540/60=90m

1x4=90x4

4s=360m

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Answer:

first option

Step-by-step explanation:

In a parallelogram consecutive angle are supplementary, sum to 180, so

2a + 122 = 180 ( subtract 122 from both sides )

2a = 58 ( divide both sides by 2 )

a = 29

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In a parallelogram the opposite sides are congruent, then

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5 0
3 years ago
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alex41 [277]

Answer:

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3 years ago
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Help plz and thx for your help
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9514 1404 393

Answer:

  Dawn has $1670 in account 1 and $1570 in account 2.

Step-by-step explanation:

Dawn can multiply the second equation by 8 and add 7 times the first equation.

  8(3/8A +7/8B) +7(A -B) = 8(2000) +7(100)

  10A = 16,700

  A = 1670

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Dawn has $1670 in account 1 and $1570 in account 2.

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<em>Check</em>

  3/8(1670) +7/8(1570) = 626.25 +1373.75 = 2000

3 0
3 years ago
A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student
yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

The information provided is as follows:

n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43

The critical value of <em>z</em> for 98% confidence level is,

z_{\alpha/2}=z_{0.02/2}=2.326

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

     =(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

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3 years ago
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Answer:

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(a * b) * c = a * (b * c)

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3 years ago
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