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3 years ago

All of the following have the same unit price except _____.

Mathematics

2 answers:

son4ous [18]3 years ago

8 0

Answer:

2 1/2 for $6.25

Step-by-step explanation:

Hello

the unit price is the price that you pay for a unit of product, you must find each unit price and compare

the unit price is given by

$Unit\ price = \frac{amount\ paid}{number\ of\ products} \\$

Step 1

find each unit price

1).2 for $5.08

$Unit\ price = \frac{amount paid}{number of products} \\Unit\ price = \frac{ 5.08}{2}\\\\Unit\ price =2.54$

unit price =2.54 $

2).5 for $12.70

$Unit\ price = \frac{amount paid}{number of products} \\Unit\ price = \frac{ 12.7}{5}\\\\Unit\ price =2.54$

unit price =2.54 $

3) 1/2 for $1.27 2

$Unit\ price = \frac{amount paid}{number of products} \\Unit\ price = \frac{ 1.27}{0.5}\\\\Unit\ price =2.54$

unit price =2.54 $

4) 2 1/2 for $6.25

$Unit\ price = \frac{amount paid}{number of products} \\Unit\ price = \frac{ 6.25}{2.5}\\\\Unit\ price =2.5$

unit price =2.5 $

Step 2

define

Now, we can define that the unit price of 2 1/2 for $6.25 is different to the others.

2.5 ≠ 2.54

Have a great day

Aliun [14]3 years ago

5 0

2 1/2 for $6.25 is the answer

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Samira and Sonia each have a bag containing 20 sweets. In each bag, there are 5 red, 6 green and 9 yellow sweets.

Volgvan

Step-by-step explanation:

there must be some typos here. but I try to address the right issues.

there is nothing under (a). so, really nothing to answer.

(b)

there are 20 sweets in her bag. 9 if the 20 are yellow.

so, the probability is the ratio of desired possibilities vs. total possibilities.

that is, tada! 9/20

that is the probabilty for Samira to pick a yellow sweet.

[1] (i)

again, 20 sweets to start with.

6 are green.

when she picks the first one, her probability to pick a green one is 6/20 = 3/10 = 0.3

and now, under the assumption that this came true, she picks another sweet.

this time she had only 19 left, and 5 of them are green.

so, this probabilty is 5/19

now both events need to happen for the case we are discussing. there is no overlapping, no ors, ifs and buts. it is just the product of both probabilities.

3/10 × 5/19 = 15/190 = 3/38

I think that is what the description asks for.

(ii)

that probability is

first selection is red and second is not red +

first selection is green and second is not green +

first selection is yellow and second is not yellow

so,

red and not red

5/20 = 1/4 red

15/19 not red (there are still 15 sweets of other colors in the bag, but again now only 19 total).

red and not red = 1/4 × 15/19 = 15/76 = 0.1974

green and not green

6/20 = 3/10 = green

14/19 = not green

green and not green = 3/10 × 14/19 = 52/190 = 26/95 =

= 0.2737

yellow and not yellow

9/20 = yellow

11/19 = not yellow

yellow and not yellow = 9/20 × 11/19 = 99/380 = 0.2606

so, now assuming up all 3 possibilities ("or" = sum) gives us the general possibility of selecting two different colors

= 0.7316

8 0

3 years ago

Suppose that insurance companies did a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle

nirvana33 [79]

Answer:

i) $ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392$

ii) $Lower = 0.8-0.0392= 0.7608$

$Upper = 0.8 +0.0392= 0.8392$

Step-by-step explanation:

Notation and definitions

$X=320$ number of people that claimed always buckle up.

$n=400$ random sample taken

$\hat p=\frac{320}{400}=0.8$ estimated proportion of people that claimed always buckle up

$p$ true population proportion of people that claimed always buckle up

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

Part i

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: