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3 years ago

One hundred nautical miles equals about 185 kilometers. To the nearest kilometer, how far in kilometers is 290 nautical miles?

Mathematics

2 answers:

nikitadnepr [17]3 years ago

6 0

The correct answer is 537 km

FromTheMoon [43]3 years ago

4 0

The correct answer is 537 :-)

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how do you use absolute value to find the distance between two points that have the same X-coordinates but different Y-coordinat

sdas [7]

So since x coordinates are same, we just have a horizontal distance to find (no fancy slanted lines)

so what you do is
|y1-y2|=distance

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3 years ago

thank you for hurrying here is the very last one Explain how you would find the perimeter of this triangle.

Lena [83]

Step-by-step explanation:

Use the method of TOA CAH SOH.

$tan(51.3) = \frac{opp}{adj} \\ \tan(51.3) = \frac{u}{4} \\ u = 4 \tan(51.3)$

$cos(51.3) = \frac{adj}{hypo} \\ \cos(51.3) = \frac{4}{v} \\ v \cos(51.3) = 4 \\ v = \frac{4}{ \cos(51.3) }$

Perimeter of triangle = Sum of all sides =

$u \: + \: v + 4$

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3 years ago

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Paul [167]

Answer:

bout 7

Step-by-step explanation:

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3 years ago

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I make $40,000 per year. I have been offered a promotion and a $5,000 raise per year. My average federal income tax rate will go

Paha777 [63]

Answer:

Yes

Step-by-step explanation:

40000 x .10 is 4000 in tax. Total after tax would be 36,000

45,000 × .15 is 6750. Total after tax would be 38,250 which is more

6 0

3 years ago

Suppose a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%. A cell phon

leva [86]

Answer:

We conclude that the actual percentage of households is equal to 30%.

Step-by-step explanation:

We are given that a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%.

Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Let p = proportion of households that have three cell phones NOT known.

So, Null Hypothesis, $H_0$ : p = 30% {means that the actual percentage of households is equal to 30%}

Alternate Hypothesis, $H_A$ : p $\neq$ 30% {means that the actual percentage of households different from 30%}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of households having three cell phones = $\frac{43}{150}$ = 0.29

n = sample of households = 150

So, the test statistics = $\frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }$

= -0.27

The value of z test statistic is -0.27.

Also, P-value of the test statistics is given by;

P-value = P(Z < -0.27) = 1 - P(Z $\leq$ 0.27)

= 1 - 0.6064 = 0.3936

Now, at 1% significance level the z table gives critical value of -2.58 and 2.58 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual percentage of households is equal to 30%.

4 0

3 years ago