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3 years ago

7

nadia typically types 32 words per minute, but her rate may vary by as much as 5 words per minute. let x be the actual rate at w

hich nadia types. what is the range of rates, in words per minute , that nadia could type

1 answer:

e-lub [12.9K]3 years ago

7 0

Answer:

trust me bro

Step-by-step explanation:

trustt meeee

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A triangle has side lengths of 200 units and 300 units. Write a compound inequality for the range of the possible lengths for th

OLga [1]

Answer:

100<x<500

Step-by-step explanation:

The third side cannot exceed the others added up so x is less than 500, due to the laws of triangles.

If you subtract 300-200, you get 100, so x has to be greater than 100.

x is greater than 100 but less than 500

Therefore the answer is...

100<X<500

The laws of triangles state that a triangle has to be greater than the other two edges subtracted, and less than the other two edges added.

I hope this helps!!!

7 0

3 years ago

For every 8 cups that Maria makes, she uses 1 1/4 cups of blueberries . If she uses 5 cups of blueberries, how many fruit cups c

Akimi4 [234]

Answer:

40

Step-by-step explanation:

3 0

3 years ago

What's bigger 5/6 or 1/2

elena-14-01-66 [18.8K]

5/6 = 0.833
1/2 = 0.5

so u see, 5/6 is bigger

7 0

4 years ago

Read 2 more answers

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

Two step equations from khan academy 3(q-7)= 27

Goryan [66]

3(q-7) = 27

q - 7 = 27/3

q - 7 = 9

q = 9 + 7 = 16

5 0

3 years ago