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3 years ago

Help me with number 7 for 10 points!!!

Mathematics

2 answers:

vodka [1.7K]3 years ago

5 0

Mode=87
Mean=77.5
Median=84
range =54

Ket [755]3 years ago

3 0

Remember this rhyme:

Hey Diddle Diddle, the medians the middle:

(Answer for your question: Median is 80%)

You add and divide for the Mean:

( Answer for your question: Mean is 93)

The mode is the one that appears the most:

( Answer for your question: Mode is 87%)

And the range is the difference between.:

( Answer for your question: Range is 54.)

I hope this helps! :)

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A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

olga_2 [115]

Answer:

1) 288.8 km due North

2) 144.9 km due East

3) 323.1 km

4) 207°

Step-by-step explanation:

Bearing: The angle (in degrees) measured clockwise from north.

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Cosine rule

$c^2=a^2+b^2-2ab \cos C$

where a, b and c are the sides and C is the angle opposite side c

-----------------------------------------------------------------------------------------------

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

Question 1

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

$\implies \sf \cos(75^{\circ})=\dfrac{y}{150}$

$\implies \sf y=150\cos(75^{\circ})$

$\implies \sf y=38.92285677$

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

Question 2

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

$\implies \sf \sin(75^{\circ})=\dfrac{x}{150}$

$\implies \sf x=150\sin(75^{\circ})$

$\implies \sf x=144.8888739$

Therefore, the ship is now 144.9 km due East.

Question 3

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

$\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)$

$\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}$

$\implies \sf d=323.1275729$

Therefore, the ship is 323.1 km from its starting point.

Question 4

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)$

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)$

$\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}$

$\implies \sf Bearing=207^{\circ}$

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

8 0

2 years ago

Read 2 more answers

Kenji is traveling in Europe and renting a car. He is used to thinking of gasoline amounts in gallons, but in Europe it is sold

finlep [7]

10.3 gallons

Good luck!!!!!

7 0

3 years ago

A high school chorus has $1000 in its school account at the beginning of the year. They are putting on a fall concert to raise m

vodomira [7]

There are several information's already given in the question. Based on those information's the answer can be easily deduced.
The equation given in the question is
T = 10x + 1000
When the ticket rate is increased to $15, the equation becaomes
T = 15x + 1000
4000 = 15x + 1000
15x = 3000
x = 200 tickets
From the above deduction, it can be concluded that the correct option among all the options that are given in the question is the third option or option "C".

3 0

3 years ago

PLEASE HELP,, MARKING BRAINLIEST!! Write the scale factor of the polygon?

Sidana [21]

Answer:

3/2

Step-by-step explanation:

just take the corresponding side lengths and divide em. you get 3/2 for everything

8 0

3 years ago

The desired percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is 5.5. To test whether the true average

Reil [10]

Answer:

a) Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

|t| = |-3.33| = 3.33

c) P - value lies between in these intervals

0.001 < P < 0.005

Step-by-step explanation:

Step( i ):-

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

Level of significance ∝ = 0.01

Step(ii):-

The test statistic

$t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }$

On calculation , we get

t = -3.33

|t| = |-3.33| = 3.33

Step(iii):-

P - value

The degrees of freedom γ = n-1 = 16-1 =15

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

Condition(i)

P - value < ∝ then reject H₀

Condition(ii)

P - value > ∝ then Accept H₀

we observe that 0.001 < P < 0.005

P- value < 0.01

we rejected H₀

(or)

The tabulated value = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

Conclusion:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

4 0

3 years ago