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2 years ago

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You were told that the 1st, 2nd and 3rd quartiles of female students' weight at a major university are 125 lbs, 138lbs, and 147l

bs. what percentage of students weigh more than 138lbs?

1 answer:

inna [77]2 years ago

3 0

Quartiles divide a dataset into 4 groups. The first quartile is the 25th percentile and has 25% of the values below it. The seond quartile, on the other hand, is the median or the 50th percentile with 50% of values below it; meanwhile, the third quartile has 75% of values below it.

Since we know that 138 lbs is the second quartile, then we can say that it is the median of the dataset. Therefore, there will be 50% of the sutdents who will weigh more than 138 lbs.

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Which of the following accurately lists all of the discontinuities for the graph below?

Murljashka [212]

Answer:

jump discontinuity at x = 0; point discontinuities at x = –2 and x = 8

Step-by-step explanation:

From the graph we can see that there is a whole in the graph at x=-2.

This is referred to as a point discontinuity.

Similarly, there is point discontinuity at x=8.

We can see that both one sided limits at these points are equal but the function is not defined at these points.

At x=0, there is a jump discontinuity. Both one-sided limits exist but are not equal.

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2 years ago

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Dan is watching The Birds in his backyard. Of the birds he watches, 9. Of them,or 45%, are sparrows. How many birds are in his b

ra1l [238]

If 9 of the sparrows is 45% of the birds in the backyard, 9/x = 45/100.

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This means that there are 20 birds in the backyard. You can check this by dividing 9/20 which equals 45%

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2 years ago

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Tanzania [10]

Answer:1,4,5,2

Step-by-step explanation:

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2 years ago

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

Viktor [21]

Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

$\Rightarrow Ae^{-36k}=46$ .......(2)

Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

$\Rightarrow -18k= ln1-ln2$

$\Rightarrow k= \frac{ln2 }{18}$

Putting the value k in equation (1)

$Ae^{-18\frac{ln2}{18}}=92$

$\Rightarrow A e^{ln2^{-1}}=92$

$\Rightarrow A.2^{-1}=92$

$\Rightarrow \frac{A}{2}=92$

$\Rightarrow A= 92 \times 2$

⇒A= 184.

Therefore $k= \frac{ln2 }{18}$ , A=184

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2 years ago

Suppose babies born in a large hospital have a mean weight of 3316 grams, and a standard deviation of 324 grams. If 83 babies ar

Anuta_ua [19.1K]

Answer: 0.129

Step-by-step explanation:

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Now, the probability that the mean weight of the sample babies would differ from the population mean by greater than 54 grams will be :

$P(|\mu-\overline{X}|>54)=1-P(\dfrac{-54}{\dfrac{324}{\sqrt{83}}}$

hence, the required probability = 0.129

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2 years ago