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2 years ago

Please help me on this now.

Mathematics

1 answer:

alexandr1967 [171]2 years ago

3 0

Answer:

the frist one is 2 the second on is 3.5 and the third one is 3 and the last on is 2.9

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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

5 0

2 years ago

Is it likely that the next 50 Sunday customers will spend an average of at least $40? Explain. Choose the correct answer below

Brrunno [24]

Answer:

B. Yes it is likely. The probability that the next 50 Sunday customers will spend an average of at least $40 is P=0.0023.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of $32 and a standard deviation of $20.</em>

We have to assume certain conditions to calculate the probability that the next 50 Sunday customers will spend an average of at least $40.

First, this 50 purchases are representative of the total purchases made in the store (the same as saying it is a random sample).

Second, the 10% condition: the 50 sales represent less than 10% of all purchases.

Third, the sample of 50 sales is large enough to make an approximation to the normal distribution.

If all these conditions are met, we can approximate the probabiltity that the next 50 Sunday customers will spend an average of at least $40.

We have a sampling distribution, with mean 32 (equal to the population mean) and standard deviation:

$\sigma_M=\dfrac{20}{\sqrt{50}}=\dfrac{20}{7.07}=2.83$

Then, we calculate the z-score

$z=\dfrac{X-\mu_M}{\sigma_M}=\dfrac{40-32}{2.83}=\frac{8}{2.83} =2.83$

The probabilty can be calculated then as:

$P(X_{50}>40)=P(z>2.83)=0.0023$

5 0

2 years ago

What is the answer to −(3x−9)

frutty [35]

Answer:

-3x + 9

Explanation:

Because these two integers that is - and - is said to be added and write the sign of of greater or bigger number.

3 0

2 years ago

Y=4x. Y=40. What does x equal

egoroff_w [7]

Answer:

Your answer is 10!

Y = 4x

(40) = 4x

40 = 4(10)

4 0

2 years ago

Read 2 more answers

Can 403/72 be reduced?

tresset_1 [31]

Answer:

yes but it would be an incomplete fraction

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

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