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3 years ago

Need the answer please, soon as possible

Mathematics

1 answer:

qaws [65]3 years ago

4 0

9514 1404 393

Answer:

(d) 27.4%

Step-by-step explanation:

The desired percentage is ...

(juniors for Kato)/(total juniors) × 100%

= 129/(129 +194 +147) × 100%

= (129/470) × 100% ≈ 27.4%

About 27.4% of juniors voted for Kato.

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If the price of a certain stock is $66.50 per share and its earnings are $4.75 per share, what is the ratio of the stock’s price

Blababa [14]

Answer:

I may be wrong but I think it's 14

5 0

2 years ago

Hannah put $2,000 in a savings

Elden [556K]

Answer:2000/180=11.11 p

Step-by-step explanation:

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6 0

3 years ago

If natalie and her friends decide to rent 4 lanes at reguler cost for a party ten people need to rent shoes and 4 people are mem

taurus [48]

The question is missing a tabular data. So, it is attached below.

Answer:

The total cost for the party is $74.50.

Step-by-step explanation:

Given:

Lanes rented at regular cost = 4

Cost of 1 lane rented at regular cost = $9.75

Cost of 1 lane rented for members = $7.50

Cost of 1 shoe rental at regular cost(non members) = $3.95

Cost of 1 shoe rental for members = $2.95

Since, lanes are rented at regular cost, we use unit rate at regular cost

So, cost of 4 lanes rented = $4\times 9.75= \$ 39$

Now, out of 10 people who rented shoes, 4 are members. So, the number of non-members is given as:

Non members who rented shoes = 10 - 4 = 6

So, 4 members and 6 nonmembers rented shoes.

So, cost of 6 non members renting shoes = $6\times 3.95=\$ 23.70$

Cost of 4 members renting shoes = $4\times 2.95=\$ 11.80$

Total cost for the party is the sum of all the costs. This gives,

= 39 + 11.80 + 23.70

= 50.80 + 23.70

= $74.50

Therefore, the total cost for the party is $74.50.

5 0

2 years ago

Add together 2350mm 3.42cm 11/2m and 1.45m

Artyom0805 [142]

9.3342meters because you convert 2350mm into m which is 2.35m, + and also convert 3.42cm to m which is 0.0342 then + 1.45m +11/2m= 9.3342m

6 0

3 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

2 years ago