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3 years ago

Seventh grade

Mathematics

1 answer:

irina [24]3 years ago

8 0

Answer:

median of seventh grade = 16.5

median of ninth grade = 16.5

Step-by-step explanation:

median = middle number, there are 18 numbers in total of each set. Since its even the middle number is kinda 16 and 17, but you can only choose one so the middle of 16 and 17 is 16.5

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olga2289 [7]

Answer:

1)x=2. 2)x=5/2

Step-by-step explanation:

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2 years ago

Use synthetic division to solve (x4 – 1) ÷ (x – 1). What is the quotient?

postnew [5]

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Step-by-step explanation:

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2 years ago

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A company is considering a new manufacturing process. It knows that the rate of savings (in dollars per year) from the process w

Ainat [17]

Answer:

i). $\$ 7500$

ii). $\$ 67500$

Step-by-step explanation:

Given in the question-

Saving rate is s(t)= 3000(t+2)

We know that savings in the 1st year can be calculated as

$\int_{0}^{1}3000(t+2).dt$

$3000\left [ \frac{t^{2}}{2}+2t \right ]_0^1$

$3000\left [ \frac{1}{2}+2 \right ]$

= $\$ 7500$

So savings in the first 5 years can be calculated as

$\int_{0}^{5}3000(t+2).dt$

$3000\left [ \frac{t^{2}}{2}+2t \right ]_0^5$

$3000\left [ \frac{25}{2}+5 \right ]$

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3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

5 0

3 years ago

I also need help with this quick!

Jlenok [28]

Answer:

I think its 5

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2 years ago

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