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2 years ago

I need help ASAP I’ll mark brainliest help please #1

Mathematics

1 answer:

Gnesinka [82]2 years ago

8 0

Answer:

Step-by-step explanation:

8-5=3, 1-(-7)=8, 3^2+8^2=73, and the square root of 73 is 8.5

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Orlov [11]

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poizon [28]

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3 years ago

If 2x/3 = 8/4, then x=

velikii [3]

Answer:

first cross multiply

2x×4=3×8

8x=24

then, divide both sides by "8" so x could be alone

x=24/8

x=3

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Step-by-step explanation:

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2 years ago

Qhich are the solutions of the quadratic equation x^2=9x+6

Lera25 [3.4K]

x^2-9x-6=0

Use the quadratic formula to get (9+sqrt(105))/2 or (9-sqrt(105))/2

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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