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3 years ago

PLZ HELP:

Mathematics

1 answer:

UkoKoshka [18]3 years ago

8 0

1. In each graph, the two lines are parallel to each other.

2. Each set of equations have the same slope.

3. Equations that have the same slope as each other will always be parallel to each other.

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Y = (x + 1)(x - 2) Distribution

stepan [7]

(x+1)2-
=it is meant to be (x+1)2
=2(x) + 2(1)
= 2x +2
I think

7 0

2 years ago

Read 2 more answers

Francis works at Carlos Bakery and is making cookie trays. She has 48 chocolate chip cookies, 64 rainbow cookies, and 120 oatmea

amm1812

The number of cookies and trays are illustrations of greatest common factors.

The number of trays is 8
6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

The given parameters are:

$\mathbf{Chocolate\ chip=48}$

$\mathbf{Rainbow=64}$

$\mathbf{Oatmeal=120}$

(a) The number of trays

To do this, we simply calculate the greatest common factor of 48, 64 and 120

Factorize the numbers, as follows:

$\mathbf{48 = 2 \times 2 \times 2 \times 2 \times 3}$

$\mathbf{64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

$\mathbf{120 = 2 \times 2 \times 2 \times 3 \times 5}$

So, the GCF is:

$\mathbf{GCF= 2 \times 2 \times 2}$

$\mathbf{GCF= 8}$

Hence, the number of tray is 8

(b) The number of each type of cookie

We have

$\mathbf{Chocolate\ chip=48}$

$\mathbf{Rainbow=64}$

$\mathbf{Oatmeal=120}$

Divide each cookie by the number of trays

So, we have:

$\mathbf{Chocolate\ chip = \frac{48}{8} = 6}$

$\mathbf{Rainbow = \frac{64}{8} = 8}$

$\mathbf{Oatmeal = \frac{150}{8} = 15}$

Hence, 6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

Read more about greatest common factors at:

brainly.com/question/11221202

4 0

2 years ago

Among cases of heart pacemaker malfunctions, were found to be caused by firmware, which is software programmed into the device.

alex41 [277]

Complete question is;

Among 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in three different pacemakers randomly selected from this batch of 8834 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?

Answer:

P(All three are not caused by firmware) = 83.84%

Probability that the entire batch will be accepted = 0.8384

Step-by-step explanation:

We are told that out of the 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware.

Thus,

Cases not caused by firmware = 8834 - 504 = 8330

So, probability of the first case not being affected by firmware is;

P(first case not caused by firmware) = 8330/8834

Also,

Probability of second case not being affected by firmware is given as;

P(second case not caused by firmware|first case not affected by firmware) = 8329/8833

Similarly,

Probability of third case not being affected by firmware is given as;

P(third case not caused by firmware|first and second not caused by firmware) = 8328/8832

Now, looking at the 3 Probabilities gotten, it is obvious that the events are not independent because the probability of occurence of one event depends on the probability of occurence of the other event.

Thus, we will make use of the general multiplication rule which is;

P(A & B) = P(B) × P(A|B)

Thus;

P(All three not caused by firmware) = P(first case not caused by firmware) × P(second case not caused by firmware|first case not affected by firmware) × P(third case not caused by firmware|first and second not caused by firmware)

Plugging in the relevant values, we have;

P(All three not caused by firmware) = (8330/8834) × (8329/8833) × (8328/8832)

P(All three are not caused by firmware) = 0.83840506679 ≈ 83.84%

6 0

3 years ago

What fraction of 11 and 1/6 in decimal form

cupoosta [38]

11.16 repeat only the six is repeating

6 0

3 years ago

Read 2 more answers

Each team in the softball league plays each of the other teams exactly once. If

Aleks04 [339]

Each team in the softball league plays each of the other teams exactly once. For every game, there is 2 team playing. The order is not important because A vs B is same as B vs A
So you just need to makes a combination of 2 that have a result of 21. If there is t number of teams, the number of matches would be:
tC2 = t!/2!(t-2)! = 21
t! / (t-2)! = 21 *2
(t)(t-1)= 42
t^2 -t -42=0
(t-7)(t+6)
t=7 ; t=-6

Excluding the minus result, you got 7 teams.

6 0

3 years ago