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2 years ago

26 is what percent of 4.16?

Mathematics

1 answer:

Anna007 [38]2 years ago

3 0

Answer:

26 is 6.25% of 4.16.

Hope this helped

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20+4*4 what the answer

schepotkina [342]

Answer: 36

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3 years ago

Read 2 more answers

Find the circum center of a triangle whose vertices are (-2, -1) , (-1, 6 ) and (2, 7).

lina2011 [118]

Answer:

(-1/3; 4)

Step-by-step explanation:

(-2, -1) , (-1, 6 ) and (2, 7).

x=(-2-1+2)/3= -1/3

y=(-1+6+7)/3=12/3=4

=> (-1/3; 4)

7 0

3 years ago

The diameter of a circle is 6 units. What is the radius of the circle?

KengaRu [80]

Answer:

Step-by-step explanation:

The radius is half of the diameter so 6/2 is 3

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2 years ago

If x-3 is a factor of x3+4x2+kx-30.Find the value of k.

Bogdan [553]

Answer:

k=-11

Step-by-step explanation:

$x^{3}+4x^{2}+kx-30$

if $x-3$ is a factor,

$(3)^{3}+4(3)^{2}+k(3)-30=0\\27+4*9+3k-30=0\\27+36+3k-30=0\\33+3k=0\\3k=-33\\k=-11$

3 0

3 years ago

For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df

rjkz [21]

Answer:

a) For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

b) For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

3 0

3 years ago