Y

Phantasy [73]

8^2-6^2=28
the square root of 28 is 5.29
5.29×2=10.58
x=10.58

7 0

4 years ago

Joe has collections of 35 DVD movies. He received eight of them as gifts. Jobot the rest of his movies over three years. If he b

Yuliya22 [10]

Hello!

Let's subtract the eight DVDs Joe received as gifts, because he didn't buy them on his own.

35 - 8 = 27

Now, divide the remaining CDs by the 3 years he collected his DVDs in.

27 ÷ 3 = 9

A N S W E R:

Joe bought 9 DVDs last year.

Good day!

8 0

3 years ago

What is the quadratic equation whose roots are -3, -1 and has a leading coefficient of 2 with X to represent the variable?

dexar [7]

$f(x)=2(x+3)(x+1)\\
f(x)=2(x^2+x+3x+3)\\
f(x)=2(x^2+4x+3)\\
f(x)=2x^2+8x+6$

3 0

3 years ago

How many different ways are there to choose a subset of the set {1,2,3,4,5,6} so that the product of the members of the subset i

zvonat [6]

You have to pick at least one even factor from the set to make an even product.

There are 3 even numbers to choose from, and we can pick up to 3 additional odd numbers.

For example, if we pick out 1 even number and 2 odd numbers, this can be done in

$\dbinom 31 \dbinom 32 = 3\cdot3 = 9$

ways. If we pick out 3 even numbers and 0 odd numbers, this can be done in

$\dbinom 33 \dbinom 30 = 1\cdot1 = 1$

way.

The total count is then the sum of all possible selections with at least 1 even number and between 0 and 3 odd numbers.

$\displaystyle \sum_{e=1}^3 \binom 3e \sum_{o=0}^3 \binom 3o = 2^3 \sum_{e=1}^3 \binom3e = 8 \left(\sum_{e=0}^3 \binom3e - \binom30\right) = 8(2^3 - 1) = \boxed{56}$

where we use the binomial identity

$\displaystyle \sum_{k=0}^n \binom nk = \sum_{k=0}^n \binom nk 1^{n-k} 1^k = (1+1)^n = 2^n$

3 0

1 year ago

Suppose that each observation in a random sample of 100 fatal bicycle accidents in 2015 was classified according to the day of t

liraira [26]

Answer:

The calculated χ² = 0.57 does not fall in the critical region χ² ≥ 12.59 so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0: proportion of fatal bicycle accidents in 2015 was the same for all days of the week

against the claim

Ha: proportion of fatal bicycle accidents in 2015 was not the same for all days of the week

2) the significance level alpha is set at 0.05

3) the test statistic under H0 is

χ²= ∑ (ni - npi)²/ npi

which has an approximate chi square distribution with ( n-1)=7-1= 6 d.f

4) The critical region is χ² ≥ χ² (0.05)6 = 12.59

5) Calculations:

χ²= ∑ (16- 14.28)²/14.28 + (12- 14.28)²/14.28 + (12- 14.28)²/14.28 + (13- 14.28)²/14.28 + (14- 14.28)²/14.28 + (15- 14.28)²/14.28 + (18- 14.28)²/14.28

χ²= 1/14.28 [ 2.938+ 5.1984 +5.1984+1.6384+0.0784 +1.6384+13.84]

χ²= 1/14.28[8.1364]

χ²= 0.569= 0.57

6) Conclusion:

The calculated χ² = 0.57 does not fall in the critical region χ² ≥ 12.59 so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

b.<u> It is r</u>easonable to conclude that the proportion of fatal bicycle accidents in 2015 was the same for all days of the week

6 0

3 years ago