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hammer [34]
3 years ago
14

Assume that 1400 births are randomly selected and 698 of the births are girls. Use subjective judgment to describe the number of

girls as significantly​ high, significantly​ low, or neither significantly low nor significantly high.
Mathematics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

neither significantly low nor significantly high.

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

Since there are a total of 1400 births we can easily calculate that half of that number is 700. Since the total number of girls born are 698 we can see that it is basically half of the total number of births. Therefore I can subjectively state that this number is neither significantly low nor significantly higher. There are almost an equal number of girls being born than there are boys.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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Allison can run 3.1 miles in 30 minutes. Which proportion could be used to find
Kobotan [32]

Answer:

3.1/30

Step-by-step explanation:

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4 0
3 years ago
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A house on the market was valued at $245,000. After several years, the value increased by 9%. By how much did the house's value
aniked [119]

We are given that a house had a value of $245000 and its value was increased by 9%. To determine how much is this value increase in dollars we need to multiply the value of the house by 9/100. We get this:

245000\times\frac{9}{100}=22050

Therefore, the increase in dollars is $22050. Now, to determine the current value we need to add the increase to the previous value, we get:

245000+22050=267050

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6 0
1 year ago
Lago enriquillo , Dominican Republic is at 151 ft below sea level. The highest mountain in the Dominican Republic is pico Duarte
riadik2000 [5.3K]

Answer:

10,013

Step-by-step explanation:

5 0
3 years ago
I need help on this please help!
Licemer1 [7]

Answer:

(b)0.56

(c)0.38

Step-by-step explanation:

(a)

P(Ben Pass) =0.8

Therefore: P(Ben fails)=1-0.8 =0.2

P(Tom Pass) =0.7

Therefore: P(Tom fails)=1-0.7 =0.3

See attached for the completed tree diagram

(b)Probability that both will pass

P(both will pass)=P(Ben pass and Tom pass)

=P(Ben pass) X P(Tom pass)

=0.8 X 0.7

=0.56

(c)The probability that only one of them will pass

Since either Tom or Ben can pass, we have:

P(only one of them will pass)

=P(Ben pass and Tom fails OR Ben Fails and Tom Pass)

=P(Ben pass and Tom fails)+P(Ben Fails and Tom Pass)

=(0.8 X 0.3) + (0.2 X 0.7)

=0.24 + 0.14

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8 0
3 years ago
Suppose f(x)= 4x/3+5 and g(x) = x/3 +1. solve for x when f(x)=g(x)
yuradex [85]
\bf \begin{cases}
f(x)=\cfrac{4x}{3}+5\\\\
g(x)=\cfrac{x}{3}+1
\end{cases}\qquad f(x)=g(x)\implies \cfrac{4x}{3}+5=\cfrac{x}{3}+1

solve for "x" then
4 0
3 years ago
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