What is the solution to the system of equations? Use the substitution method.

Dan invests £8000 into his bank account.

Serhud [2]

If 8000 dollars multiplied by 2% is 160 dollars and 160 dollars multiplied by 3 years gives you the answer of 480 dollars

6 0

3 years ago

Read 2 more answers

What are the real zeros of the function g(x) = x3 + 2x2 − x − 2?

Nat2105 [25]

Upon a slight rearrangement this problem gets a lot simpler to see.

x^3-x+2x^2-2=0 now factor 1st and 2nd pair of terms...

x(x^2-1)+2(x^2-1)=0

(x+2)(x^2-1)=0 now the second factor is a "difference of square" of the form:

(a^2-b^2) which always factors to (a+b)(a-b), in this case:

(x+2)(x+1)(x-1)=0

So g(x) has three real zero when x={-2, -1, 1}

4 0

2 years ago

-12<-6x please help

pishuonlain [190]

-12 < -6x

-12/-6 > x divide both sides by -6

12/6 > x two negatives makes a positive

2 > x simplify

x < 2 switch sides

hope this helps, God bless!

6 0

3 years ago

Write the recursive formula for the sequence a^n =2-5(n-1)

Zolol [24]

The formula is 2”6 hope I helped

3 0

2 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago