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3 years ago

Here are the first five terms of Fibonacci sequence.

Mathematics

1 answer:

Pepsi [2]3 years ago

5 0

Answer:

a- 32,52

b-20+32=52

Step-by-step explanation:

I think there is only one 4

as following 4,8,12,20

4+8=12

8+12=20

20+12=32

20+32=52

52+32=84

84+52=136

136+84=220

136+220=356

so 4,8,12,20,32,52,84,136,220,356

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A girl makes 12 foul shots for every 8 that she misses. How many shots did she make if she shot 125 foul shots

Gala2k [10]

Answer:

Step-by-step explanation:

set up a proportion: 12/20 = x/125

so x=75 :)

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3 years ago

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Jalen bought 4 bottles of water, 2 bags of cashews, and a fruit salad. The cost of the fruit salad was $5.

Ede4ka [16]

Answer:2

Step-by-step explanation: 2x2=4 therefore 4 is double 2

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3 years ago

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Soloha48 [4]

Answers:

1a) y=-10/3x+90

1b) 20

1c) -18

2a) 2.8

2b) How much the heights of five basketball players vary from the average height.

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Explanations:

1a) The trend line is linear, so we just need to find the slope and y-intercept to find an equation for it. Our y-intercept is (0,90), or 90, and our slope is -10/3. Our equation is now y=-10/3x+90.

1b) To find when x=21, we plug 21 into our equation where the x is. Now we do the math.

y=-10/3(21)+90 (plug in)

y=-70+90 (simplify by multiplying -10/3 by 21)

y=20 (simplify by adding -70 to 90)

Therefore, we can predict that when x is 21, y is 20.

1c) To find when y=150, we plug 150 into our equation where the y is. Now we do some more math.

150=-10/3x+90 (plug in)

60=-10/3x (subtract 90 from both sides

-18=x (divide both sides by -10/3)

Therefore, we can predict that when y is 150, x is -18.

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

2a) The mean absolute deviation (or MAD for short) of a data set is calculated as such:

Step 1) Find the mean (average) by finding the sum of the data values, then dividing the sum of the data values by the number of data values. In this case, we have the numbers 65, 58, 64, 61, and 67, which add up to 315. The data set has 5 numbers, so we divide our sum of 315 by 5 to get 63. Therefore, our mean is 63.

Step 2) Find the absolute value of the distance between each data value and the mean. In this case, we find out how far away each data value is from 63, our mean. To do this, we subtract 63 from each number.

65-63=2

58-63=-5

64-63=1

61-63=-2

67-63=4

Some of these values are negative, but we're using absolute value so they all become positive. We now have a new set of values: 2, 5, 1, 2, and 4.

Step 3) Finally, we calculate the mean of our new set of values. In this case, we will add up 2, 5, 1, 2, and 4 to get 14 and divide by 5 to get our MAD of 2.8. Therefore, the MAD (and the answer to problem 2a) is 2.8.

2b) Now we just find out what the MAD means in this context. The MAD always is a measure of variance in a data set. In this context, it's describing how much the heights (in inches) of five people on a basketball team vary from the average height.

Hope this helps!

3 0

3 years ago

Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.

jek_recluse [69]

Answer:

$\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

Step-by-step explanation:

Let $X$ be a set of all students in your class. The set $X$ is the domain. Denote

$C(x) - ' \text{$x $ has a cat}'\\D(x) - ' \text{$x$ has a dog}'\\F(x) - ' \text{$x$ has a ferret}'$

$\mathbf{a)}$

Consider the statement 'A student in your class has a cat, a dog, and a ferret'. This means that $\exists x \in X$ so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)$

$\mathbf{b)}$

Consider the statement 'All students in your class have a cat, a dog, or a ferret.' This means that $\forall x \in X$ at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)$

$\mathbf{c)}$

Consider the statement 'Some student in your class has a cat and a ferret, but not a dog.' This means that $\exists x \in X$ so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)$

$\mathbf{d)}$

Consider the statement 'No student in your class has a cat, a dog, and a ferret..' This means that $\forall x \in X$ none of the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

$\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)$

$\mathbf{e)}$

Consider the statement ' For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.'

This means that for each of the statements C, F and D there is an element from the domain $X$ so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

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4 years ago

What is a unit rate?

qwelly [4]

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3 years ago

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