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3 years ago

Erin buys 4 packs of potatoes costing £2 each, 2 packs of carrots for £3 per pack, and 5 turnips costing 70p each.

1 answer:

8 0

She would get £2.50

4 times 2 is 8
2 times 3 is 6
0.7 times 5 is 3.5
Add them together
Minus it from 20
It equals £2.50

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Complete the statement. Explain your reasoning.If the measure of ∠2 = 48∘, then the measure of ∠3 = _______.

const2013 [10]

Answer:

132 degrees

Step-by-step explanation:

the angles of measure 2 and 3 have to add up to 180 degrees (because that's the angle of a straight line, which 2 and 3 make together)

180-48=132

6 0

3 years ago

Please help ill give brainliest and 5 stars

grandymaker [24]

Answer:

that looks like it should be C

Step-by-step explanation:

7 0

3 years ago

.4^3Grafting, the uniting of the stem of one plant with the stem or root of another, is widely used commercially to grow the ste

11111nata11111 [884]

Answer:

0.784 = 78.4% probability that there will be at least one failed graft in the next three done

Step-by-step explanation:

To solve this question, we need to understand the binomial probability distribution and conditional probability.

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.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: No failed grafts in the first seven

Event B: At least one fail in the next three.

Intersection of events A and B:

Since the probability of a graft failling is independent of other grafts, we have that:

$P(A \cap B) = P(A)*P(B)$

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)*P(B)}{P(A)} = P(B)$

So we just have to find the probability of one fail in three trials.

Three trials means that $n = 3$ .

The probability is

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{3,0}.(0.4)^{0}.(0.6)^{3} = 0.216$

Then

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.216 = 0.784$

0.784 = 78.4% probability that there will be at least one failed graft in the next three done

3 0

3 years ago

Need HELP ASAP PLEASEEEEEEE

timofeeve [1]

Answer:B

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

g 5. (5 points) Next, let’s turn our attention to the vector space P2, which is the set of polynomial with degree at most 2, tog

almond37 [142]

Answer:

No, those vectors do not span P₂

Step-by-step explanation:

At first, notice we have 4 vectors from a vector space of dimension 3 ({1,x,x²} is the standard base of P₂) so we are sure they are linearly dependent (because there are more than 3 vectors) i.e. it´s possible to write one of them in terms of other in the same set:

p4=p1-p2

$(2x^{2} -x+1) - (x+3) = 2x^{2}-2x-2$

Now, we conclude the span {p1,p2,p3,p4} = {p1,p2,p3}

If we want to know the subspace that this span gets, we work with the matrix of the vectors made as follows:

$A=\left[\begin{array}{ccc}1&3&5\\-1&1&-1\\2&0&4\end{array}\right]$

The row-reduction of this matrix show us the basis of the same row-subspace even in the case that some of them are linearly dependent:

$A\ is\ row--equivalent\ to:\left[\begin{array}{ccc}1&0&\frac{1}{2}\\0&1&-\frac{3}{2}}\\0&0&0\end{array}\right]$

This shows that the span {p1,p2,p3}={(1+0.5x²),(x-1.5x²)}

Looking in the previous equality, the span {p1,p2,p3} is generated by two linearly independent vectors, but P₂ needs 3 linearly independent vectors to be generated.

The answer is that {p1,p2,p3,p4}⊂P₂.

4 0

4 years ago