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enot [183]
3 years ago
9

Convert to higher terms. 3/15 ?/30

Mathematics
2 answers:
Arada [10]3 years ago
6 0
The answer is 6/30 :)
kirill [66]3 years ago
4 0

Answer:

I would assume its 6/30. and the fact your asking this dumb of a question means you need to study first grade math instead of being on brainly.

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Sati [7]
10 3/5 is the correct answer
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4 years ago
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Find the length of a line joining the point ( 4,3 ) and origin .​
GalinKa [24]

Answer:

i think you have to count till you get to 4 or 3 then the remaining you plus with the 3

5 0
3 years ago
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F(x) = 2x - 3 and f(x) = f'(x), find the value of x.​
Salsk061 [2.6K]

Answer:

x=1

Step-by-step explanation:

According to your typed question,

f(x)=2x-3

Let f(x)=y

y=2x-3

Now,

Interchanging the positions of x and y

x=2y-3

x+3=2y

x+3/2=y

f'(x)=x+3/2

Then,

f(x)=f'(x)

2x-3=x+3/2

2(2x-3)=x+3

4x-3=x+3

4x-x=3+3

3x=3

x=3/3

x=1

According to your image question,

f(x)=x/2x-3

f(x)=f'(x)

Now,

Let y=f(x)

y=x/2x-3

y(2x-3)=x

2xy-3=x

2xy-x=3

x(2y-1)=3

2y-1=3x

2y=3x+1

y=3x+1/2

f'(x)=3x+1/2

Then,

f(x)=f'(x)

x/2x-3=3x+1/2

2x=6xsq+2x-9x+3

2x=6xsq+7x+3

solve for x ok

8 0
3 years ago
A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team’s previously te
puteri [66]

Answer:

a. z = 2.00

Step-by-step explanation:

Hello!

The study variable is "Points per game of a high school team"

The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)

The hypothesis is:

H₀: μ ≤ 99

H₁: μ > 99

α: 0.01

There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.

The statistic value under the null hypothesis is:

Z= X[bar] - μ  = 101 - 99 = 2

σ/√n 6/√36

I don't have σ, but since this is an approximation I can use the value of S instead.

I hope it helps!

7 0
3 years ago
You come home from Brian’s Orchard with a big brown bag of apples: 23 Granny Smiths, 14 Honey Crisp and 31 Red Delicious. What i
garik1379 [7]

Answer:

a. Probability of Pulling one of each = 0.03175

b. Probability of Pulling 4 Honey Crisp = 0.001797

Probability of 2 G.Smith = 0.1144

Probability of 1 Red Delicious = 0.4559

Step-by-step explanation:

Given

Granny Smiths = 23

Honey Crisp = 14

Red Delicious = 31

Required

- Probability of Pulling out one of each

- Probability of Pulling out 4 Honey Crisp; 2 Granny Smiths; 1 Red Delicious

First, the total number of apple needs to be calculated.

Total = Granny Smiths + Honey Crisp + Red Delicious

Total = 23 + 14 + 31

Total = 68

Probability of Pulling 1 of each

= P(Granny Smiths) and P(Henry Ford) and P(Red Delicious)

- Granny Smiths;

This is calculated by dividing number of Granny Smiths apples by total number of apples.

Probability = 23/68

Similarly,

Probability of Pulling Honey Crisp= Number of Honey Crisp divided by total

Probability = 14/68

Probability = 7/34

Probability of Pulling Red Delicious = Number of Red Delicious divided by total

Probability = 31/68

So, Probability of Pulling 1 of each = 23/68 * 7/34 * 31/68

Probability = 4991/157216

Probability = 0.03175

Probability of Pulling out 4 Honey Crisp;

= P(Honey) * P(Honey) * P(Honey) * P(Honey)

= (P(Honey))⁴

= (7/34)⁴

= 2401/1336336

= 0.001797

Probability of Pulling 2 Granny Smiths;

= P(Granny) * P(Granny)

= (P(Granny))²

= (23/68)²

= 529/4624

= 0.1144

Probability of 1 Red Delicious

= number of red delicious divided by total

= 31/68

= 0.4559

3 0
3 years ago
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