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Elanso [62]
3 years ago
14

It takes has a 20 minutes to walk to school . it takes Andre 80% as long to walk to school how long dose it take Andre to walk t

o school
Mathematics
1 answer:
Blizzard [7]3 years ago
7 0
You have to find 80% of 20.
Since 80% is equal to 0.8, you multiply 20 by 0.8 to get 16
You might be interested in
11111111111111111111111
Nat2105 [25]

Answer:

1) Zero based on (-16·t - 2) is t = -1/8 second

2) Zero based on (t - 1) is t = 1 second

Step-by-step explanation:

The given functions representing the height of the beach ball the child throws as a function of time are;

y = (-16·t - 2)·(t - 1) and y = -16·t² + 14·t + 2

We note that (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

Therefore, the function representing the height of the beachball, 'y', is y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The zeros of a function are the values of the variables, 'x', of the function that makes the value of the function, f(x), equal to zero

In the function of the question, we have;

y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The above equation can be written as follows;

y = (-16·t - 2) × (t - 1)

Therefore, 'y' equals zero when either (-16·t - 2) = 0 or (t - 1) = 0

1) The zero based on (-16·t - 2) = 0, is given as follows;

(-16·t - 2) = 0

∴ t = 2/(-16) = -1/8

t = -1/8 second

The zero based on (-16·t - 2) is t = -1/8 second

2) The zero based on (t - 1) = 0, is given as follows;

(t - 1) = 0

∴ t = 1 second

The zero based on (t - 1) is t = 1 second

4 0
3 years ago
Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introvert
Setler [38]

Answer:

P(x \ge 5) = 1.000 ---- At least 5 from marketing departments are extroverts

P(x=15) = 0.013 ---- All from marketing departments are extroverts

P(x = 0) = 0.002 ---------- None from computer programmers are introverts

Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

P(x) = ^nC_x * p^x * (1 - p)^{n-x}

(a):\ P(x \ge 5)

n = 15 --- marketing personnel

p = 75\% --- proportion that are extroverts

Using the complement rule, we have:

P(x \ge 5) = 1 - P(x < 5)

So, we have:

P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}

P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}

P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}

P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153

P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103

So, we have:

P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103

P(x < 5) = 0.00011922283

Recall that:

P(x \ge 5) = 1 - P(x < 5)

P(x \ge 5) = 1 - 0.00011922283

P(x \ge 5) = 0.9998

P(x \ge 5) = 1.000 --- approximated

(b)\ P(x = 15)

n = 15 --- marketing personnel

p = 75\% --- proportion that are extroverts

So, we have:

P(x) = ^nC_x * p^x * (1 - p)^{n-x}

P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}

P(x=15) = 1 * 0.75^{15} * (0.25)^{0

P(x=15) = 0.013

(c)\ P(x = 0)

n=5 ---------- computer programmers

p = 70\% --- proportion that are introverts

So, we have:

P(x) = ^nC_x * p^x * (1 - p)^{n-x}

P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}

P(x = 0) = 1 * 1 * (0.30)^5

P(x = 0) = 0.002

3 0
2 years ago
It’s possible to build a triangle with side lengths of 5,5, and 10
nikitadnepr [17]
No.  The two lengths 5 and 5 add up to 10; you'd have two sides of length 5 each placed end to end, equaling 10.  That'd be represented by segments on a straight line, not a triangle.

Hope you've drawn a picture, so as to understand this better.  ;)

7 0
3 years ago
Read 2 more answers
Using only whole numbers Nikki wrote as many multiplication equations as she could with 12 as the product what where her equatio
erastovalidia [21]
1 times 12,2 times 6,and 3times 4
4 0
3 years ago
Prove the trig identity: (tanx/1+secx)+(1+secx/tanx)=2cscx
lakkis [162]
=[(sinx/cosx)/(1+1/cosx)] + [(1+1/cosx)/(sinx/cosx)]

=[(sinx/cosx)/(cosx+1/cosx)]+[(cosx+1/cosx)/(sinx/cosx)]

= [sinx/(cosx+1)] + [(cosx+1)/sinx]

= [sin^2x+(cosx+1)^2] / [sinx (cosx+1)]

= [2+2cosx] / [sinx(cosx+1)]

=[2(cosx+1)] / [sinx (cosx+1)]

= 2/sinx

= 2 cscx

(I think this will be helpful for you. if you can see the picture, it has more detail in it.)

7 0
3 years ago
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