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2 years ago

9

Craig has £13.40 he sees this offer in a restaurant :main courses £8.90 each.Buy one main course and get the second half price.

Can he afford to buy two main courses?Show your working

1 answer:

kap26 [50]2 years ago

8 0

Answer:

Yes

Step-by-step explanation:

Full price for first course: £8.90

Half price: £4.45

One and one half times

full price is then: £13.35

Since this is less than Craig's £13.40, he can just barely afford to buy two main courses at these prices.

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Which expression(s) have a greatest common factor (GCF) of 3xy2 with 42xy4

Aleks [24]

Answer:

3xy2

Step-by-step explanation:

GCF is the largest factors which are common in two or more numbers.

We find the GCF of two or more number number by breaking each number into their factor and then observe which factors are common in the group and then we multiple the common number to get the GCF.

Lets understand this by an example

GCF of 24 and 36

factors of 12 = 2*2*2*3 , factors of 36 = 2*2*3*3

hence we see that 2,2 and 3 are common in 12 and 36 hence

GCF will be 2*2*3 = 12

Coming back to problem

given

Expression 1 is 3xy^2

factor of 3xy^2 = 3*x*y*y

Expression 2 is 42xy4

factor of 42xy4 = 2*3*7*x*y*y*y*y

common factor between the two expressions are

3, x, y , y

hence GCF will be = 3*x*y*y =3xy2 Answer

3 0

2 years ago

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in n

devlian [24]

Answer:

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so $\mu = 49.9, \sigma = 3.74$ .

What is the length in cm of the longest 15% of Atlantic cod in this area?

We have to find the value of X for the value of Z that has a pvalue of 0.85.

Looking at the zscore table, we have that Z = 1.04 has a pvalue of 0.8508. So, we have to find the value of X when $Z = 1.04, \mu = 49.9, \sigma = 3.74$ .

So

$Z = \frac{X - \mu}{\sigma}$

$1.04 = \frac{X - 49.9}{3.74}$

$X - 49.9 = 3.8896$

$X = 53.7896$

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

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2 years ago

Jorge solves the equation 4x-(x+2)+6

Burka [1]

Answer:

Step-by-step explanation:

2x+6

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Callie

4 0

2 years ago

Read 2 more answers

Just the question about sandy))<br><br>A & B

hammer [34]

A.) 50.45m + 60 = 57.95m

b.) 50.45m + 60 = 57.95m
60 = 57.95m - 50.45m
60 = 7.50m
60 / 7.50 = m
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7 0

2 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

2 years ago