Please help with test offering big points pt 3

3/5 divide by 2/3 help it's for my homework

gregori [183]

Answer:

.9

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the solution to the following system of equations?

Semmy [17]

Answer:

(3,2)

Step-by-step explanation:

The system is

x + y = 5

x -y = 1

Add the equations to eliminate y

2x = 6----->x=3

Substitute this value in any equation

3+y = 5----->y=2

The solution is (3,2)

3 0

3 years ago

Y = -2x + 2 y = 5x + 9

WITCHER [35]

X=-1 y=4 hope it helps

8 0

2 years ago

Not sure what to do here? Naomi has 50 red beads and white beads. The number of red beads is 1 more than 6 times the number of w

lorasvet [3.4K]

R + w = 50
r = 6w + 1

now we sub ..
6w + 1 + w = 50
7w + 1 = 50
7w = 50 - 1
7w = 49
w = 49/7
w = 7 <== white beads

r = 6w + 1
r = 6(7) + 1
r = 42 + 1
r = 43 <=== red beads

7 0

3 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