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3 years ago

Solid fats are more likely to raise blood cholesterol levels than

Mathematics

2 answers:

alina1380 [7]3 years ago

8 0

Answer:

Step-by-step explanation:

We would determine the mean and standard deviation for the stick and liquid fats.

Mean = sum of items/number of I teams

For the stick margarine,

n = 6

Mean = (25.5 + 26.7 + 26.5 + 26.6 + 26.3 + 26.4)/6 = 26.3

Standard deviation, s1 = √(summation(x - mean)^2/n

Summation(x - mean)^2 = (25.5 - 26.3)^2 + (26.7 - 26.3)^2 + (26.5 - 26.3)^2 + (26.6 - 26.3)^2 + (26.3 - 26.3)^2 + (26.4 - 26.3)^2 = 0.94

s1 = √0.94/6 = 0.41

For liquid margarine,

n = 6

Mean = (16.5 + 17.1 + 17.5 + 17.3 + 17.2 + 16.7)/6 = 17.1

Summation(x - mean)^2 = (16.5 - 17.1)^2 + (17.1 - 17.1)^2 + (17.5 - 17.1)^2 + (17.3 - 17.1)^2 + (17.2 - 17.1)^2 + (16.7 - 17.1)^2 = 0.73

s2 = √0.73/6 = 0.4

Since we know the sample standard deviation and the sample sizes are small, we would determine the t test statistic by applying the formula,

(x1 - x2)/√(s1²/n1 + s2²/n2)

x1 and x2 are the sample means

Therefore,

t = (26.3 - 17.1)/√(0.41²/6 + 0.4²/6)

t = 39.34

Sholpan [36]3 years ago

4 0

Answer:

Step-by-step explanation:

mean of stick= 26.33

mean of liquid= 17.05

standard deviation of stick= 0.394

standard deviation of liquid= 0.345

Here we are comparing two samples so we will use t-test

t-statistic= (mean1-mean2)/√(SD1²/(N1-1) + SD2²/(N2-1))

= 39.6

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Which property applies only to addition and subtraction

ad-work [718]

Addition you add and in subtraction you subtract

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2 years ago

What is the LCM of 4,8,14?

Tamiku [17]

The LCM of 4,8,14 is 56

M of 4 are 4,8,12,16,20,24,28,32,36,40,44,48,52,56
M of 8 are 8,16,32,48,56,64
M of 14 are 14,28,42,56

I hope this helps you

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3 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

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3 years ago

How do you subtract fractions

olya-2409 [2.1K]

Answer:

make the denominators ( bottom number ) the same and whatever u do to that do to the top

Step-by-step explanation:

e.g ½ - ⅓

make the common denominator 6

to get from 2 to 6 u times by 3 so it wud be 3/6

and then to get from 3 to 6 u times by 2 so do the same to the numerator (top number) so u get 2/6

3/6 take away 2/6 is 1/6

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2 years ago

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What decimal is equivalent to negative 3/8

Fittoniya [83]

Answer:

-6/16

Step-by-step explanation:

<u>Step 1: Find equivalent to -3/8</u>

(-3*2) / (8*2)

Answer: -6/16

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2 years ago

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