Harry buys a TV priced at £1200 plus 20% VAT.

4 The number of laps Jessica runs around the track is proportional to the time she

iragen [17]

Answer:

A. 24 lap/hour.

B. r = 24h

Step-by-step explanation:

Let r be the number laps.

Let h be the time.

From the question given above, the number of laps (r) Jessica runs around the track is proportional to the time (h) she spend running. This can be written as:

r & h

r = Kh

Where:

r is the number of laps.

K is the constant of proportionality.

h is the time.

A. Determination of the constant of proportionality.

With the above formula, we can calculate the value of the constant of proportionality, K as follow:

Number of laps (r) = 12 laps

Time (h) = ½ hour

Constant of proportionality (K) =.?

r = Kh

12 = K × ½

12 = K/2

Cross multiply

K = 12 × 2

K = 24 lap/hour

Therefore, constant of proportionality is

24 lap/hour.

B. Equation for relationship between the number of lap and time. This is illustrated below:

r = Kh

K = 24

r = 24h

Therefore, the equation for relationship between the number of lap and time is r = 24h

7 0

3 years ago

Plz help I will give brainliest<br> What is the value of x in this problem<br> 97x+14=13.8

sammy [17]

X= -1/485 (-0.00206)

8 0

3 years ago

Two boys decided to buy a computer. The second boy had 5/6 of the money the first had. The first boy had 7/8 of the price of the

UNO [17]

Answer:

The price of the computer is $1,152

Step-by-step explanation:

Let the price of the computer be $x

The first boy boy had 7/8 of this amount, the amount he has is thus 7/8 × $x = 7x/8

The second boy had 5/6 of what the first boy had. The amount of money he has is thus 5/6 × 7x/8 = 35x/48

Now, the addition of what they have is $696 more than what they need to pay. This amount is x+ 696

Mathematically, this can be represented as;

7x/8 + 35x/48 = x + 696

(42x + 35x)/48 = (x + 696)

77x/48 = (x + 696)

77x = 48(x + 696)

77x = 48x + 33408

77x - 48x = 33408

29x = 33, 408

x = 33,408/29

x = $1,152

7 0

3 years ago

According to the University of Nevada Center for Logistics Management, 6% of all mer-

Fudgin [204]

Answer:

a) The point estimate of the proportion of items returned for the population of

sales transactions at the Houston store = 12/80 = 0.15

b) The 95% confidence interval for the proportion of returns at the Houston store = [0.0718 < p < 0.2282].

c) Yes.

We set an hypothesis and construct a test statistics. The test statistics result gives us:

Z calculated = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns for the nation as a whole.

Step-by-step explanation:

a) Point estimate of the proportion = number of returned items/ total items sold = 12/80 = 0.15.

b) By formula of confident interval:

CI(95%) = p ± Z* $\sqrt{\frac{p*(1-p)}{n} } = 0.15 \pm 1.96 *\sqrt{\frac{0.15*(1-0.15)}{80} }$ ,

CI(95%) = [0.0718 < p < 0.2282]

c) The hypothesis:

$H_{0}$ : The proportion of returns at the Houston store is not significantly different from the returns for the nation as a whole.

$H_{a}$ : The proportion of returns at the Houston store is significantly different from the returns for the nation as a whole.

The test statistics:

Z = $\frac{\hat{p} - p_{0}}{\sqrt{\frac{p*(1-p)}{n} }}$ , where $p_{0}$ is the proportion of nation returns.

Z calculated = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns for the nation as a whole.

6 0

3 years ago

Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin

NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)

Step-by-step explanation:

1) Previous concepts

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Confidence interval

Assuming that $\bar X =5.5$ and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=1086-1=1085$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that $t_{\alpha/2}=2.33$

Now we have everything in order to replace into formula (1):

$5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139$

$5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861$

So on this case the 95% confidence interval would be given by (5.139;5.861) We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

5 0

3 years ago