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3 years ago

I'm trying to find the standard form for 8.97x10 to rh negative 1. Can you helpe?

1 answer:

jeka943 years ago

7 0

10000 = 1 x 10424327 = 2.4327 x 104 1000 = 1 x 1037354 = 7.354 x 103 100 = 1 x 102482 = 4.82 x 102 10 = 1 x 10189 = 8.9 x 101 (not usually done) 1 = 100
 1/10 = 0.1 = 1 x 10-10.32 = 3.2 x 10-1 (not usually done) 1/100 = 0.01 = 1 x 10-20.053 = 5.3 x 10-2 1/1000 = 0.001 = 1 x 10-30.0078 = 7.8 x 10-3 1/10000 = 0.0001 = 1 x 10-40.00044 = 4.4 x 10-4 . Look at this.

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3 years ago

Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people

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Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

$p = 0.85, n = 6$ .

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

$P(X < 6) + P(X = 6) = 1$

$P(X < 6) = 1 - P(X = 6)$

In which:

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

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229$

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$

$P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993$

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236$

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

$P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960$

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

$P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376$

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

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3 years ago

Nick has 3 shirts: a white one, a black one, and a blue one. He also has two pairs of pants, one blue and one tan. What is the p

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english has 26 different letters. how many consecutive words have to be selected in a book to have at least 5 that start with th

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The words need not be English language words 142506

What is permutation and combination?

A set of elements can be divided into subsets in two different ways: combination and permutation.

The components of the subset may be arranged in any order when combined. The components of the subset are listed in a permutation in a certain order.

If the word has 5 different letters then it is just 26-choose-5 or (26/5).

If the word has 4 different letters then it is just 26-choose-4 or (26/4)

have 4 options to choose which letter to duplicate twice.

This gives = 4*(26/4)

If the word has 3 different letters then it is just 26-choose-3 or (26/3).

you duplicate one of the three letters 3 times - 3 options for that, or you choose 2 letters and duplicate each one twice - again 3 options. In total this gives = 6*(26/3).

If the word has 2 different letters then it is just 26-choose-2 or (26/2).

The sets of 2 different letters and duplicate either one of the letters 4 times - 2 options; or duplicate one letter 2 times, and the other 3 times - again 2 options. In total this is = 4*(26/2).

The word has just 1 distinct letter you have = (26/1).

we get :

N = (26/5)+ 4*(26/4) + 6*(26/3) + 4*(26/2) + (26/1)

= 142506

The words need not be English language words = 142506

Learn more about permutation and combination

brainly.com/question/28065038

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1 year ago