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3 years ago

Lesson 8 homework practice solve percent problems

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

I can help you out!!

lesantik [10]3 years ago

5 0

Use slader.com they give you all the answer on there , look up the subject , the book you need and then you have all the answers there.

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A sandwich shop sells sandwiches for $5.00 each (tax included). On Friday afternoon, the shop received a total of $75.00 for the

matrenka [14]

Answer:

75/5=x

Step-by-step explanation:

take the total sales divided by the unit price to give you the number of units

4 0

3 years ago

In a batch of 440 water purifiers, 8 were found to be defective. What is the probability that a water purifier chosen at random

LiRa [457]

Answer:

1.8% or 1.818181%

Step-by-step explanation:

If in a batch of 440 water purifiers, we find 8 to be defective.

Then the probability of the next purifier being defective is 8/440

In percentage = 8/440 * 100

=> 8/44 * 10

=> 8/22 * 5

=> 40/22

=> 20 / 11

20/11 is roughly 1.8181818

Rounding it to the nearest tenth, we get 1.8%

If my answer helped, kindly mark me as the brainliest!!

Thank You!!

3 0

3 years ago

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

If y = 5x − 3, which of the following sets represents possible inputs and outputs of the function, represented as ordered pairs?

andre [41]

U would solve this by subbing in ur answer choices to see which ones make the equation true.

ur answer is : (-1,-8) , (0,-3), (3,12)

4 0

3 years ago

I need help with this question!<br> Experts only please <br><br> Will mark brainliest

Naily [24]

3x + 2y(100)
yes he was correct
2 cups are added each time so 2 times 100 is 200 add the 3 you have 203.
i know i’m not an expert but i wanted to help :)

4 0

3 years ago