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bekas [8.4K]
3 years ago
12

Two boxes are similar, the length of the side of the small box is 4cm, the length of the side of the big box is 16cm. The volume

of the big box is 320cm3 what is the volume of the small box?
Mathematics
1 answer:
MariettaO [177]3 years ago
6 0

Answer:

80cm^3

Step-by-step explanation:

16/4 = 4

This means that the scale factor = 4

320cm^3 = Volume of BB

BB/4 = Volume of SB

320/4 = 80

Therefore, the volume of the small box = 80cm^3.

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Explorers on a planet with a thin atmosphere want to measure the acceleration of gravity at its surface. So they use a spring gu
Murljashka [212]

Answer:

The value of g_s is g_s=\frac{30}{21} \:\frac{m}{s^2}.

Step-by-step explanation:

We know that the position function is given by

s(t)=30t-\frac{g_s}{2} t^2

Velocity is defined as the rate of change of position. Therefore,

v(t)=\frac{d}{dt}(s(t))

So, the velocity function is

v(t)=\frac{d}{dt}(s(t))\\\\v(t)=\frac{d}{dt}(30t-\frac{g_s}{2} t^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(30t\right)-\frac{d}{dt}\left(\frac{g_s}{2}t^2\right)\\\\v(t)=30-g_st

When a body reaches a vertical velocity of zero, this is the maximum height of the body and then gravity will take over and accelerate the object downward. Thus,

v(t)=30-g_st=0\\\\g_s=\frac{30}{t}

We know that the ball bearing reaches its maximum height 21 seconds after being launched (t = 21 s).

So, the value of g_s is g_s=\frac{30}{21} \:\frac{m}{s^2}.

5 0
4 years ago
I will vote u brainliest if u answer so pls do
Lera25 [3.4K]

Answer: For this, you have to calculate the Unit value,

Unit value of 1rst one,  4.99/6 = 0.83

Unit value of 2nd one, 7.99/10 = 0.79

As, 2nd one is cheaper, so it is better :))

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
3x+7=13 (x=-2;x=2;x=5)
AnnZ [28]
X=2 because 3 times 2 is 6 and then you add 6 and 7 together and get 13.
5 0
3 years ago
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The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and
kykrilka [37]

Answer:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent

The p-value is 0 .001909. The result is significant at p < 0.05

Part b:

40 > 8.5

35> 7.5

25> 4

Step-by-step explanation:

1) Let the null and alternative hypothesis as

H0: the quality of management and the reputation of the company are independent

against the claim

Ha: the quality of management and the reputation of the company are dependent

2) The significance level alpha is set at 0.05

3) The test statistic under H0 is

χ²= ∑ (o - e)²/ e where O is the observed and e is the expected frequency

which has an approximate chi square distribution with ( 3-1) (3-1)=  4 d.f

4) Computations:

Under H0 ,

Observed       Expected E              χ²= ∑(O-e)²/e

40                      35.00                          0.71

25                      24.50                         0.01

5                         10.50                         2.88  

35                      40.00                         0.62

35                      28.0                          1.75

10                       12.00                           0.33  

25                      25.00                             0.00

10                        17.50                              3.21

<u>15                       7.50                                 7.50  </u>

<u>∑                                                               17.0281</u>

     

     

Column Totals 100 70 30   200  (Grand Total)

5) The critical region is χ² ≥ χ² (0.05)2 = 9.49

6) Conclusion:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent.

7) The p-value is 0 .001909. The result is significant at p < 0.05

The p- values tells that the variables are dependent.

Part b:

If we take the excellent row total = 70 and compare it with the excellent column total= 100

If we take the good row total = 70 and compare it with the good column total= 80

If we take the fair row total = 50 and compare it with the fair column total= 30

The two attributes are said to be associated if

Thus we see that ( where (A)(B) are row and columns totals and AB are the cell contents)

AB> (A)(B)/N  

40 > 1700/200

40 > 8.5

35> 1500/200

35> 7.5

25> 800/200

25> 4

and so on.

Hence they are positively associated

8 0
3 years ago
Use the distributive property to simplify the expression<br> 5(3 +x) – 2
insens350 [35]

5(3 +x) – 2 = (5+3)*(5+x)

                       (8-2)*(5x-2)

                         6*5x-2

this Is what I came up with.

4 0
4 years ago
Read 2 more answers
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