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3 years ago

Help please asap!

Mathematics

2 answers:

WARRIOR [948]3 years ago

6 0

Answer:

I think the answer is "How much pineapple juice is needed to make 5 pitchers of punch?"

Step-by-step explanation:

Komok [63]3 years ago

5 0

I believe it's the second one, from 5 cups of juice, since you would divide 5 by 3 1/2

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Find the area of the trapezoid

lidiya [134]

area = ( top rule + bottom rule ) × height / 2

area = ( 2 + 3 ) × 4 / 2

area = 5 × 4 / 2

area = 20 / 2

area = 10

4 0

2 years ago

Read 2 more answers

Trains Two trains, Train A and Train B, weigh a total of 360 tons. Train A is heavier than Train B. The difference of their weig

Sergio [31]

Since both trains in total weigh 360 tons and train A weighs more than train be, it means that train B is the rest of the weight when you subtract train A from the total amount. Difference means subtraction so just solve this: 360-272 to get train B and you already know that train A is 272 tons. Please mark branliest!

8 0

3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$

$Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)$

$(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}$

$Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}$

$Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}$

$n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2$

$n = (16*(\frac{1.2}{(270-264)}))^2$

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0

3 years ago

What is the diameter of a circle whose radius is 6.45 meters

Vikentia [17]

12.9
Diameter is just the radius times 2

6 0

3 years ago

The sun produces 3.9 ⋅ 10 to the 33th power ergs of radiant energy per second. How many ergs of radiant energy does the sun prod

arlik [135]

So the sun produces 3.9x10³³/ second

Using this ratio (3.9x10³³/ second) we can use the porpotions to calculate time taken (T) to produce 1.55x10⁷.

$\frac{3.9(10^3) ergs}{1 second} = \frac{1.55(10^7)ergs}{T}$

Then isolate T and find the time

$\frac{T3.9(10^3)ergs}{1 second} = {1.55(10^7)ergs$
$T3.9(10^3)ergs = ({1.55(10^7)ergs)seconds$
$T = \frac{(1.55(10^7)ergs)seconds}{3.9(10^3)ergs}$
$T = \frac{(1.55(10^7)seconds}{3.9(10^3)}$
$T = \frac{1.55}{3.9}(10^4)seconds$
$T = 0.397(10^4)seconds$
$T = 3.97(10^-^1)(10^4)seconds$
$T = 3.97(10^3)seconds$

Thus it took 3.97x10³ seconds for 1.55x10⁷. of ergs to be produced
Hope that helped and if it did, a Brainliest rating would be much appreciated

6 0

3 years ago