(k^8+8k^2+10k+21) / (k+7)

Lori wants to distribute 35 peaches equally into baskets.She will use more than 1 but fewer than 10 baskets.How many baskets doe

Brums [2.3K]

Answer:

Step-by-step explanation: you have to divide the peaches into 7 separate baskets and each of those 7 baskets there would be 5 peaches.

7 * 5=35

4 0

3 years ago

In order to solve for the variable in the equation 2 (x + 3) + 5 x = 3 (2 x minus 1), Jaleesa begins by applying the distributiv

avanturin [10]

Answer: 7 x + 6 = 6 x minus 3

Step-by-step explanation:

2 (x + 3) + 5 x = 3 (2 x minus 1)
Now you distribute.
2(x) + 2(3) = 2x+6
And 3(2x) minus 3(1) = 6x minus 3
2x+6+5x = 6x minus 3
Then combine like terms
2x +5x=7x and there are no other like terms on either side of the equation.
7x+6= 6x minus 3

8 0

3 years ago

Read 2 more answers

Triclosan is a compound often added to products such as soaps, lotions, and toothpaste. It is antimicrobial, so for that reason,

miv72 [106K]

Answer:

The appropriate alternative hypothesis is, Hₐ: p₁ - p₂ > 0.

Step-by-step explanation:

In this case it is provided that Triclosan is a compound often added to products such as soaps, lotions, and toothpaste.

Since the compound is antimicrobial its corporate producer maintained that it should help prevent staph infections among people who use products containing the compound.

But a scientist wants to test whether the compound actually is associated with higher incidence of staph infection.

That is he wants to determine whether people with a detectable level of triclosan have higher rates of carrying the staph bacteria than people without triclosan.

Denote:

Group 1: people with a detectable level of triclosan

Group 2: people without a detectable level of triclosan

To test this the hypothesis can be defined as follows:

H₀: The proportion of people with a detectable level of triclosan does not have higher rates of carrying the staph bacteria than people without triclosan, i.e. p₁ - p₂ ≤ 0.

Hₐ: The proportion of people with a detectable level of triclosan have higher rates of carrying the staph bacteria than people without triclosan, i.e. p₁ - p₂ > 0.

Thus, the appropriate alternative hypothesis is, Hₐ: p₁ - p₂ > 0.

6 0

3 years ago

An urn contains nine red and one blue balls. A second urn contains one red and five blue balls. One ball is removed from each ur

il63 [147K]

Answer:

0.362

Step-by-step explanation:

When drawing randomly from the 1st and 2nd urn, 4 case scenarios may happen:

- Red ball is drawn from the 1st urn with a probability of 9/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this case to happen is (9/10)*(1/6) = 9/60 = 3/20 or 0.15. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 5 blue)/(8 red + 1 blue + 5 blue) = 6/14 = 3/7.

- Red ball is drawn from the 1st urn with a probability of 9/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (9/10)*(5/6) = 45/60 = 3/4 or 0.75. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 4 blue)/(8 red + 1 blue + 1 red + 4 blue) = 5/14

- Blue ball is drawn from the 1st urn with a probability of 1/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (1/10)*(5/6) = 5/60 = 1/12. The probability that a ball drawn randomly from the third urn is blue given this scenario is (4 blue)/(9 red + 1 red + 4 blue) = 4/14 = 2/7

- Blue ball is drawn from the 1st urn with a probability of 1/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this event to happen is (1/10)*(1/6) = 1/60. The probability that a ball drawn randomly from the third urn is blue given this scenario is (5 blue)/(9 red + 5 blue) = 5/14.

Overall, the total probability that a ball drawn randomly from the third urn is blue is the sum of product of each scenario to happen with their respective given probability

P = 0.15(3/7) + 0.75(5/14) + (1/12)*(2/7) + (1/60)*(5/14) = 0.362

8 0

3 years ago

Find two consecutive odd integers such that seven times the first equals five times the second.

fredd [130]

Answer:

Odd integers are always one more that an even integer:

1st: 2n+1

2nd: 2n+3

Step-by-step explanation:

14n+7 = 10n+15

4n = 8

n = 2

------------

1st: 2n+1 = 5

2nd: 2n+3 = 7

7x=5*(x+2)

7x=5x+10

7x-5x=10

2x=10

x=5

8 0

3 years ago