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2 years ago

(76-x+(17+4x)(25+7x)

Mathematics

2 answers:

daser333 [38]2 years ago

8 0

I believe this is correct

Ivanshal [37]2 years ago

4 0

Answer:

Idk

Step-by-step explanation:

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Scientists discovered a large star 400,000 light years from Earth.A light-year is about 5,880,000,000,000 miles.What is the dist

Sedbober [7]

So if you want to find the distance, you would take 400,000x5,880,000,000,000=
2.35200000E+18. So your answer would be 2.35200000E+18 in scientific notation it would be 2.35200000x10^18. I hope this helps!

3 0

3 years ago

Use the quadratic formula to solve the equation.<br> x^2 – 7x – 6 = 0<br><br> Show work please.

victus00 [196]

Answer:

The standard form of quadratic equation $ax^2+bx+c = 0$ , then the values of x which are the solutions of the equation are given by:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ .

Given the equation: $x^2-7x-6=0$

Here, a=1 , b= -7 and c = -6.

then, the solution of the given equation given by:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the values of a, b and c in above we get;

$x=\frac{-(-7)\pm\sqrt{(-7)^2-4(1)(-6)}}{2(1)}$ or

$x=\frac{7\pm\sqrt{49+24}}{2} =\frac{7\pm\sqrt{73}}{2}$

then,

the solutions of the given equation are;

$x=\frac{7+\sqrt{73}}{2}$ , $\frac{7-\sqrt{73}}{2}$ .

4 0

2 years ago

Read 2 more answers

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

Alborosie

Answer:

The 80% confidence interval for the average net change is (8.596, 12.904).

Critical value t=1.638.

Step-by-step explanation:

First, we calculate the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{4}(12+7+13+11)\\\\\\M=\dfrac{43}{4}\\\\\\M=10.75\\\\\\s=\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2\\\\\\s=\dfrac{1}{3}((12-10.75)^2+(7-10.75)^2+(13-10.75)^2+(11-10.75)^2)\\\\\\s=\dfrac{20.75}{3}\\\\\\s=6.92\\\\\\$

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=10.75.

The sample size is N=4.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.63}{\sqrt{4}}=\dfrac{2.63}{2}=1.315$