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3 years ago

12

Consider the following formula used to estimate the height, h, of a seeding after w weeks since planting

2 answers:

a_sh-v [17]3 years ago

8 0

B, 5h-8 is the answer

Aleonysh [2.5K]3 years ago

5 0

H= 1/5W + 8/5
h- 8/5 = 1/5w
5h-8 = w

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In a store, a sample of 60 pens was selected from a box of 1,200 pens. Of these 60, 4 pens did not work. About how many pens cou

Lapatulllka [165]

Ok so if 1200 total pens and 60 were taken that's a ratio of 1200/60=20

out of 60pens 4 didn't work, again ratio is 4/60=1/15 so about 1/15 of the pens didn't work.

so we have 1200pens * 1/15broken ratio = 80 pens or we can say

20*4=80pens won't work from 20 groups of 60 and 4 out of 60 not working is 20*4=80 not working from total batch of 1200.

hope this helps you some! thank you!!!!!!!!!

3 0

2 years ago

An airplane takes 3 hours to travel a distance of 1680 miles with the wind. The return trip was 4 hours against wind. Find the s

nasty-shy [4]

Answer:

Speed of plane in still air: 490 mile/hour

speed of wind: 70 mile/hour

Step-by-step explanation:

still speed = x wind speed = w

with wind: speed = still + wind

(x+w) * 3 = 1680 ...a

against wind: speed = still - wind

(x-w) * 4 = 1680 ...b

b-a ... 4x-4w-3x-3w = 0

x-7w = 0

x = 7w

substitution: (7w+w)*3 = 1680 8w = 560 wind speed (w) = 70

still speed (x) = 7w = 490 mile/hour

5 0

2 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

2 years ago

12x-39≤9 AND −4x+3<−6

Elza [17]

Answer:

9/4<x≤4 or all real numbers

Step-by-step explanation:

I'm not sure what you are asking

4 0

3 years ago

Diego solved an equation by multiplying both sides of the equation by 6. Then he checked that 6 is the correct solution by subst

Studentka2010 [4]

C. on e2020 . Good luck!

6 0

2 years ago