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3 years ago

9

You put the following items into your rucksack. Ham 600 g Cheese 380 g Tomatoes 270 g How much do all the ingredients weigh in t

otal?

1 answer:

nekit [7.7K]3 years ago

5 0

The answer is 1,250 as you add all the items together

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ASAP Point A, located at (-3, -3) is translated 6 units up, resulting in the location of point A'. What other transformation wou

miss Akunina [59]

Answer:

(3,3)

Step-by-step explanation:

A reflection of this across the y-axis is (-3,3). A reflection of Point A, (-3,-3), over the x-axis is also (-3,3).

3 0

3 years ago

Sunday by 11:59pm

beks73 [17]

Answer:

The amount of markup is 88 dollars.

Step-by-step explanation:

Given that the whole sale price of bicycle is: $400

$P = \$400$

Mark up is the profit that the seller of the product makes after selling the product more in price more than its cost or buying price.

Given that mark up percentage is 22%

It can be written mathematically as:

$Mark\ up = 22\%\ of\ Price\\Mark\ up = 0.22 * P\\Mark\ up = 0.22 * 400\\= \$88$

Hence,

The amount of markup is 88 dollars.

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3 years ago

I need the answer will mark

zvonat [6]

Answer:

z to the 10th power

Step-by-step explanation:

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2 years ago

Read 2 more answers

5+ (x + 2) = <br><br> !!!!!!!!!!!!!!!!!!!!!

DiKsa [7]

Answer:

x + 7

Step-by-step explanation:

5 + (x + 2) = x + 7

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3 years ago

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You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r

kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that $p = 0.15$

Assume you obtain a random sample of 9 individuals from this population:

This means that $n = 9$

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

$P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005$

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0

2 years ago