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timama [110]
3 years ago
6

Amit’s school bus travelsat 36km/hr and Sushma’s school bus travels at 11 m/s. Whose school bus is faster ?

Mathematics
2 answers:
UNO [17]3 years ago
8 0

Answer:

This answer is not originally mine so Don give me credit

Step-by-step explanation:

Amit's school bus in m/s will be

36×1000/3600

36 × 10/36

= 10 m/s

While sushma's school bus has speed of 11 m/s

so sushma's school bus is faster than Amit's school bus

gogolik [260]3 years ago
5 0

Answer:

Amit's: 10 m/s

Sushma's 11 m/s

Step-by-step explanation:

Amit's:

36 x 1000/3600

36 x 10/36

Sushma's:

11 m/s

Sushma's school bus is faster than Amit

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Answer:

15/121 is the answer.

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forsale [732]

Answer:

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\frac{4 + 9 - 10}{15}

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hope it helps you :)

have a great day

7 0
3 years ago
Read 2 more answers
Would love some help :)
Usimov [2.4K]

Answer:so would i

Step-by-step explanation:

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3 years ago
Y varies inversely with x, y = 4 and x = 3, what is y, when x = 2B) given: y = 2x - 1 (x-3)(x+1) what is the I. VAII. HA III. X-
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SOLUTION

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\begin{gathered} y\alpha\frac{1}{x} \\  \\ y=\frac{k}{x} \\  \\ 4=\frac{k}{3} \\  \\ k=12 \end{gathered}\begin{gathered} y=\frac{12}{x} \\  \\ when\text{ x=2} \\  \\ y=\frac{12}{2} \\  \\ y=6 \end{gathered}

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1 year ago
A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop
Katyanochek1 [597]

Answer:   (0.367,\ 0.473)

Step-by-step explanation:

The confidence interval for population mean is given by :-

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

, where \hat{p} is the sample proportion, n is the sample size , z_{\alpha/2} is the critical z-value.

Given : Significance level : \alpha:1-0.99=0.01

Sample size : n= 85

Critical value : z_{\alpha/2}=2.576

Sample proportion: \hat{p}=\dfrac{36}{85}\approx0.42

Now, the  99% confidence level will be :

\hat{p}\pmz_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{\dfrac{0.42(1-0.42)}{85}}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)

Hence, the  99% confidence interval estimate of the true proportion of families who own at least one DVD player is  (0.367,\ 0.473)

3 0
3 years ago
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