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3 years ago

I'm so confused xD picture attached.

Mathematics

2 answers:

marta [7]3 years ago

8 0

The answer would be D because once you write out the equation in the problem it will look like the equation shown in B but that's not simplified in scientific notation so because there are 4 zeros in 10000 you would add 4 to the -14 and -14+4=10

dimulka [17.4K]3 years ago

4 0

B. if you think it is not correct don't put it in

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IVE BEEN TRYING TO GET THE ANSWER FOR THIS WITH A PICTURE AND KEYBOARD PLEASE HELP (It’s a picture)

serg [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

For a family with 3 children, what is the probability that they have exactly 2 boys and 1 girl?

Triss [41]

Answer:

$\frac{3}{8}$

Step-by-step explanation:

Possibilities (G for girls and B for boys):

GGG

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4 0

3 years ago

Read 2 more answers

Can anybody explain to me what is being done here ?

bija089 [108]

Answer:

So first, they remove the denominator of x times x, because multiplication cancels out division, so then you're left with

3/1 times x = 2193

3/1 just equals three, so you write it down as 3.

3x = 2193

To isolate 3x, you need to divide by 3.

x = 731

Hope this helped :)

5 0

2 years ago

Read 2 more answers

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 199 times. Here are the observed

Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28, 29, 47, 40, 22, 33.

Probability of an Event

$=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}$

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

$(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12$

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

$(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911$