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3 years ago

Which is a better deal: A can of 3 tennis balls for $3 or a box of 10 balls for $9?

Mathematics

1 answer:

Kay [80]3 years ago

6 0

Answer:

a box of 10 balls for $9?

Step-by-step explanation:

because u buy 10 tennis balls for 9 dollars and the first one is not true

becuase u buy 3 balls of tennis with 3 dollars and the last answer is right

becuase you buy 10 balls for 9 that means its 1 dollar less for 10 balls

hope it helps

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A marketing research company needs to estimate the average total compensation of CEOs in the service industry. Data were randoml

mezya [45]

SOLUTION:

Step 1 :

In this question, we are told that a marketing research company needs to estimate the average total compensation of CEOs in the service industry.

We also have that: Data were randomly collected from 38 CEOs and the 98% confidence interval was calculated to be ($2,181,260, $5,836,180).

Then, we are asked to find the margin error for the confidence interval.

Step 2:

We need to recall that:

$\text{Higher Confidence Interval, CI}_{H\text{ = }}X\text{ + }\frac{Z\sigma}{\sqrt[]{n}}$ $\text{Lower Confidence Interval , CI}_{L\text{ }}=\text{ X - }\frac{Z\sigma}{\sqrt[]{n}}$

It means that:

$\vec{}X\text{ = }\frac{CI_{H\text{ }}+CI_L}{2}$ $\text{Margin of error, }\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{CI_{H\text{ - }}CI_L}{2}$

where,

$CI_H\text{ = }$$5,836,180$\text{ }$$$$ $CI_{L\text{ }}=\text{ }2,181,260$

putting the values into the equation for the margin of error, we have that:

$\text{Margin of error,}\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{5,836,180\text{ - }2,181,260\text{ }}{2}$ $\begin{gathered} =\text{ }\frac{3654920}{2} \\ =1,\text{ 827, 460} \end{gathered}$

CONCLUSION:

The margin error for the confidence interval is 1, 827, 460

5 0

1 year ago

Let X equal the number of flips of a fair coin that are required to observe

Nataliya [291]

Answer:

(a) I attached a photo with the diagram.

(b) $f(x) = \big( \frac{1}{2} \big)^{x-1}$

(d) 4

(e) $k^2/2$

Step-by-step explanation:

(a) I attached a photo with the diagram.

(b) The easiest way to think about this part is in terms of combinatorics. Think about it like this.

To begin with, look at the three each level of the three represents a possible outcome of throwing the coin n-times. If you throw the coin 3 times at the end in total there are 8 possible outcomes. But The favorable outcomes are just 2.

1 - Your first outcome is HEADS and all the others are different except the last one.

2 - Your first outcome is TAILS and all the others are different except the last one.

Therefore the probability of the event is

$f(x) = P(X=x) = \frac{2}{2^x} = \frac{1}{2^{x-1}} = \big( \frac{1}{2} \big)^{x-1}$

(c)

P(X = 0) = 0 because it is not possible to have two consecutive tails or heads.

$E(X > 4) = 1 - P(X \leq 3) = 1 - ( P(X = 0 ) + P(X = 1) + P(X = 2) + P(X = 3))\\= 1 - ( \frac{1}{2} + \frac{1}{4} ) = \frac{1}{4}$

(d)

Remember that this is a geometric distribution therefore

$E[X] = p/(1-p)$ , in this case $p = 1/2$ so $E[X] = 1$ and

$E[X+1]^2 = ( E[X] +1 )^2 = (1+1)^2 = 2^2 = 4$

Also

(e)

This is a geometric distribution so its variance is

$Var(X) = \frac{1-p}{p^2} = 1/2 / 1/4 = 1/2$

And using properties of variance

$Var(kX - k ) = Var(kX) = k^2 var(X) = k^2 /2$

3 0

4 years ago

Find the fifth term of the arithmetic sequence in which t1 = 3 and tn = tn-1 + 4. A) 5 B) 7 C) 19 D) 23 2) Find the tenth term o

Sergeu [11.5K]

We know that

Part a) Find the fifth term of the arithmetic sequence in which t1 = 3 and tn = tn-1 + 4
t1=3
t2=t1+4----> 3+4-----> 7
t3=t2+4-----> 7+4----> 11
t4=t3+4-----> 11+4---> 15
t5=t4+4-----> 15+4---> 19

the answer Part a) is 19

Part b) Find the tenth term of the arithmetic sequence in which t1 = 2 and t4 = -10

we know that
tn=t1+(n-1)*d-----> d=[tn-t1]/(n-1)
t1=2
t4=-10
n=4
find the value of d
d=[-10-2]/(4-1)-----> d=-12/3----> d=-4
find the tenth term (t10)
t10=t1+(10-1)*(-4)----> t10=2+9*(-4)----> t10=-34

the answer Part b) is -34

Part c) Find the fifth term of the geometric sequence in which t1 = 3 and tn = 2tn-1
t1=3
t2=2*t1----> 2*3----> 6
t3=2*t2----> 2*6----> 12
t4=2*t3-----> 2*12---> 24
t2=2*t4----> 2*24----> 48

the answer Part c) is 48

3 0

4 years ago

I want the answer................

RUDIKE [14]

Answer:

AB = CD = 3√5

Step-by-step explanation:

We use Pythagoras' Theorem.

AB² = CD² = 3²+6² = 9+36 = 9·(1+4) = 9·5

=> AB = CD = √(9·5) => AB = CD = 3√5

5 0

3 years ago

Read 2 more answers

What is the factor y^2-11y+18

Nezavi [6.7K]

Answer:

(y-9)(y-2)

Step-by-step explanation:

-9 and -2 add up to -11 and when multiplied they make a positive 18

6 0

3 years ago

Read 2 more answers