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masha68 [24]
3 years ago
15

PLEASE PLEASE PLEASE HELP!!

Mathematics
1 answer:
Fittoniya [83]3 years ago
8 0
Okay, ratios!  Yah!

So heres how it works.

Lets just make sure we have the info.

So we have butter, sugar, and vanilla at a 23:8:1 ratio.  We know that this all must add up to get 96 tablespoons.

Well look at that.  There is a nice number we can use hidden in here.

Add up the ratio amounts and we have 32 tablespoons in total.  Well thats easy.  All we need to do is multiply by 3 and we get 96 tablespoons.

So, the answer would be 3 tablespoons of vanilla
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Thi sis the question
jek_recluse [69]

\text{Given that,}\\\\x^2 + \dfrac1{x^2} = 38\\\\\\(i)\\\\\left(x-\dfrac 1x \right)^2 = x^2 + \dfrac 1{x^2} - 2 \cdot x \cdot \dfrac 1x\\\\\\\implies \left(x-\dfrac 1x \right)^2 = 38 -2 = 36\\\\\implies x - \dfrac 1x = \pm\sqrt{36} = \pm 6 \\\\\\(ii)\\\\x^2 + \dfrac 1{x^2} = 38\\\\\\\implies \left(x^2 + \dfrac 1{x^2} \right)^2 = 38^2\\\\\\\implies x^4 + \dfrac 1{x^4} + 2\cdot x^2 \cdot \dfrac 1{x^2} = 1444\\\\\\\implies x^4 + \dfrac 1{x^4} + 2 = 1444\\\\\\

\implies x^4 +\dfrac 1{x^4} = 1444-2 = 1442

7 0
2 years ago
The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level. F
OlgaM077 [116]

Answer:

The Sample size is 1918.89035

Step-by-step explanation:

Consider the provided information.

It is given that 14 out of 105 samples failed.

Therefore p = 14/105 = 0.13 3... and q=1-0.133=0.867

Samples would be needed to create a 99 percent confidence interval.

Subtract the confidence level from 1, then divide by two.

\frac{(1 -0.99)}{2}=0.005

By standard normal table z=2.5758≈2.58

Calculate the sample size as:

n=\frac{z^2pq}{e^2}

Where,  e is the margin of error,

Substitute the respective values.

n=\frac{(2.58)^2(0.133)(0.867)}{(0.02)^2}=1918.89

Hence, the Sample size is 1918.89035

4 0
3 years ago
Suppose that the weight of an newborn fawn is Uniformly distributed between 2.5 and 4 kg. Suppose that a newborn fawn is randoml
Lubov Fominskaja [6]

Answer:

a) The mean is 3.25

b) The standard deviation is 0.433

c) The probability that fawn will weigh exactly 3.7 kg is 0

d) The probability that a newborn fawn will be weigh between 2.9 and 3.5 is 0.4

e) The probability that a newborn fawn will be weigh more than 3.3 is 0.4667

f) The probability that a newborn fawn will be weigh more than P(x > 2.9 | x < 3.7) is 0.6667

g) The 59th percentile is 3.385

Step-by-step explanation:

a) In order to calculate the mean we would have to make the following calculation:

mean = (4 + 2.5) / 2 = 3.25

b) In order to calculate the standard deviation we would have to make the following calculation:

standard deviation = (4 - 2.5) / √(12) = 0.433

c) P(X = 3.7) = 0

d)  In order to calculate the probability that a newborn fawn will be weigh between 2.9 and 3.5 we would have to make the following calculation:

P(2.9 < X < 3.5) = (3.5 - 2.9) / (4 - 2.5) = 0.4

e) In order to calculate the probability that a newborn fawn will be weigh more than 3.3 we would have to make the following calculation:

P(X > 3.3) = (4 - 3.3) / (4 - 2.5) = 0.4667

f) P(X > 2.9 | X < 3.7) = P(X > 2.9 and X < 3.7) / P(X < 3.7) = P(2.9 < X < 3.7) / P(X < 3.7) = [(3.7 - 2.9) / (4 - 2.5)] / [(3.7 - 2.5) / (4 - 2.5)] = 0.6667

g)  In order to calculate the 59th percentile we would have to make the following calculation:

P(X < x) = 0.59

(x - 2.5) / (4 - 2.5) = 0.59

x = 3.385

6 0
4 years ago
Find x if the average of 20, 20, 19, 13, and x is 16
fiasKO [112]
By adding 20+20+19+13+x and dividing by 5 you will receive the answer. Therefore by substituting numbers, 8 will give you the correct average.
20+20+19+13+8 = 80
80/5 = 16.
Therefore x = 8
If I have helped you out, help me out by making me the brainliest answer.
4 0
4 years ago
Read 2 more answers
Fiona is goin to jog once around the block she lives on. The block is rectangular, and it is 180 yards long and 200 feet wide. H
irakobra [83]

Answer:

Fiona will jogged 1,080 feet

Step-by-step explanation:

180 x 2 = 360 yards

360 x 3 = 1,080 feet

3 0
3 years ago
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