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Ira Lisetskai [31]
3 years ago
11

9. Jonathan typically answers 8 out of 10 of his math

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
8 0
I believe he should expect 10 wrong because 40/50 is the same as 8/10
Masja [62]3 years ago
8 0

Answer:

I would say that the answer to the question of your's is 40/50

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What's the y intercept of 4x+3y=9
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X-intercept - (9/4, 0)
Y-intercept - (0, 3)

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3 years ago
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Which box-and-whisker plot shows the high temperatures in Pittsburgh, Pennsylvania during the first two weeks of March: 33, 45,
nikitadnepr [17]
There is no picture so i am not sure what to tell u
6 0
3 years ago
Help please, will give lots of points!
Troyanec [42]

Step-by-step explanation:

roots as exponents are fractions

if there is no root number next to the radical sign its assumed that its 1/2

3√2 is 2^(1/3)

√3 is 3^(1/2)

4√3 is 3^(1/4)

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8 0
2 years ago
Find the minimum sample size needed to be 99% confident that the sample's variance is within 30% of the population's variance.
liq [111]

The Minimum sample size table is attached below

Answer:

X=173

Step-by-step explanation:

From the question we are told that:

Confidence Interval CI=99\%

Variance \sigma^2=30\%

Generally going through the table the

Minimum sample size is

X=173

7 0
3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
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