Three sixth-grade students competed in the 100-meter dash. The table shows their times.

Mathematics

1 answer:

Sergeeva-Olga [200]2 years ago

5 0

Answer:

It is 12.9, 13.1, 13.2 C.

Step-by-step explanation:

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A student concluded that the solution to the equation sqrt2x+1 +3=0 is x=4. Do you agree? Why or why not?

Katyanochek1 [597]

Just put 4 as x in the equation and solve it. The new equation is:

sqrt(2*4)+1+3 = 0

Now we can solve it.

sqrt(2*4) becomes sqrt8:

sqrt8+1+3=0. -------> sqrt8+4=0

The sqrt of 8 is about 2.83.. so 2.83+4 is:

6.83

So we have 6.83 as the answer

but 6.83 does not equal 0 so x cannot equal 4

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2 years ago

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8_murik_8 [283]

Answer:4/15

Step-by-step explanation:

Change the whole number into a fraction and divide both and you get 4/15 or get photomath with math problems works perfectly

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2 years ago

Graph the functions to find The solution. f(x)=x^2 - 2 g(x)= 2x + 1

larisa [96]

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2 years ago

Suppose GH = JK , HI = KL and <1 =

Stella [2.4K]

I think the answer is <1=34

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2 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

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2 years ago