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3 years ago

PLZ, help me with this question.

Mathematics

2 answers:

GenaCL600 [577]3 years ago

7 0

Answer:

The first one

Step-by-step explanation:

klasskru [66]3 years ago

4 0

Answer:

Impractical

Step-by-step explanation:

It's not practical to have a ship hold up for so long also new technology will arrive in the next 100 years no need to use outdated technology

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4a+6b=10 2a-4b=12 what does 12a =?

Usimov [2.4K]

If you have at least as many equations as your have unknown variables, the 'system' is solvable (unless the equations are copies of eachother).

In this case, isolate one letter and plug it in the other. I'm going for the 2a in the bottom one (it doesn't matter)

2a - 4b = 12 => (divide by two and move the b's to the other side)

a = 6 + 2b. (plug this one into the top equation)

4(6+2b) + 6b = 10 => 24+8b+6b = 10 => 24 + 14b = 10 => 14b = -14 => b=-1

a = 6 + -2 = 4.

So a=4 and b=-1.

4 0

3 years ago

?/1 What percent of 29 is 3? Round to the nearest tenth of a percent.

Marina CMI [18]

Answer:

10.34%

Step-by-step explanation:

29×x%=3

X=10.34%

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 $z_{\alpha/2}=1.96$