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2 years ago

HELP FAST ANYONE OLEASE I HAVE LIKE 2 minutes too answer !!

Mathematics

2 answers:

kow [346]2 years ago

3 0

Answer:

Step-by-step explanation:

notsponge [240]2 years ago

3 0

Answer: c

Step-by-step explanation: I JUST ANSWERED BOTH PARTs

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Jayden buys a game for $10.67 and some batteries for $5.14 he pays with a twenty dollar bill estimate how much change he should

natima [27]

Answer:

4.19

Step-by-step explanation:

If rounded to the nearest dime 4.20

nice

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3 years ago

The graph shows two linear functions, f and g. Which formula BEST represents g(x)?

Softa [21]

Answer: C

g(x) = 1/2 f (x)

Step-by-step explanation:

3 0

3 years ago

We know that a class interval must be greater than what the I formula solution is; however... What is the solution for class int

stich3 [128]

Answer:

Therefore the answer is 20.

Step-by-step explanation:

We know that

class interval = range / number of classes

But here number of classes is not given , so we use the formula

class interval = range / ( 1+ 3.322 log N)

where , range =maximum - minimum = 220-100 = 120

N= number of observations = 50

class interval = 120 / ( 1+ 3.322 * log 50) = 18.06

Rounding up to a convinient number

Thus , class intervai = 20

Therefore the answer is 20.

3 0

2 years ago

A leak in a pool causes the height of the water to decrease by 0.25 foot over 2 hours. After the leak is fixed, the height of th

lana [24]

Answer:

5 feet

Step-by-step explanation:

Add 0.25 to both sides of the equation:

4.75 +0.25 = x - 0.25 +0.25

5.00 = x

The original height of the water in the pool was 5.00 feet.

7 0

3 years ago

Read 2 more answers

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago