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2 years ago

Work out the percentage change to 2 decimal places when a price of £87.95 is decreased to £70.

Mathematics

2 answers:

Ann [662]2 years ago

8 0

Answer:

20.40%

Step-by-step explanation:

percentage change = difference divided by original x 100

Difference here: 17.95

Thus, % change = 17.95 divided by 97.95 x 100 = 20.40%

Hopr this helps.

kap26 [50]2 years ago

5 0

Answer:

20.41%

Step-by-step explanation:

87.95 - 70 = 17.95

17.95 / 87.95= 0.20409323479

0.20409323479 x 100 = 20.409323479

round it

= 20.41%

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The number of hours Layla worked each week for the first eight weeks of summer break are recorded below.

Rainbow [258]

Answer: The value of the center increases and the distribution has a smaller spread.

Step-by-step explanation:

Re - arranging the first value in ascending order , we have :

12 ,14, 16 , 19, 21 , 22 , 28 , 32

Median = 19 + 21 / 2

= 40/2

= 20

Also calculating the standard variation in order to know the spread , we need to first of all calculate the mean

Mean = 12 + 14 + 16 + 19 + 21 + 22 + 28 + 32 / 8

Mean = 164/8

Mean = 20.5

Therefore to calculate the standard deviation, we need to calculate the variance, which is

$(12-20.5)^{2}$ + $(14-20.5)^{2}$ + $(16-20.5)^{2}$ + $(19-20.5)^{2}$ + $(21-20.5)^{2}$ + $(22-20.5)^{2}$ + $(28-20.5)^{2}$ + $(32-20.5)^{2}$ / 8

Variance = 328/8

Variance = 41

Standard deviation = $\sqrt{variance}$

S.D = $\sqrt{41}$

S.D = 6.4

Also re arranging the second values , we have :

20 , 24 , 25 , 31

Median = 24 + 25 / 2

Median = 49/2

Median = 24.5

Calculating the S.D using the same format , the S.D = 3.9

Comparing the two we can conclude that the value at the center increases and the distribution has a smaller spread

8 0

2 years ago

Read 2 more answers

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

1 year ago

The general form of the equation of a circle is 7x2 + 7y2 − 28x + 42y − 35 = 0. The equation of this circle in standard form is

Ede4ka [16]

Answer:

The standard form of a circle is (x-h)^2 + (y-k)^2 = r^2 with (h,k) being the center of the circle and r being the radius. In this case the circle's equation in standard form is (x-2)^2 + (y+3)^2 = 18. Knowing this it's easy to see that the center of the circle (h,k) is (2,-3). Finally the radius is $\sqrt{18}$ or in simplified terms, 3 $\sqrt{2}$

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Von mixes tomato sauce and milk to make soup. First he puts 150 milliliters of

elixir [45]

Answer:

C, 350

Step-by-step explanation:

So we know that each dark mark on the measuring cup represents 100 mL. Using this information, we can now discover that the lighter marks are 50 mL more than a darker mark. Because the liquid is in between 300 and 400, we can take the average of these numbers to get a final answer of 350.

Hope this helps!

7 0

1 year ago

Read 2 more answers

on a sunny day liam decided to walk to his grandmother's house. After walking for a while at a rate of 3 mph, Liam realized that

inna [77]

Answer:

the answer is 10

Step-by-step explanation:

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1 year ago