If

Olenka [21]

The answer would be 150 as each ten bar is si worth 10 and 10x15=150.

8 0

3 years ago

An experienced cashier at a grocery store takes 2 seconds to scan each item and 40 seconds to process the customer's payment. 1)

grigory [225]

Answer:

The correct option is D) 43 items.

Step-by-step explanation:

As we know that total given time is ,

$Y = (60\times2) +6 =126\hspace{0.15cm} sec$ ,

Out of which cashier spends 40 seconds to process the customer's payment.

So the remaining time = (126 - 40) = 86 secs.

And the remaining time is used to scan the item and so that we can calculate here the no items scan is = (86 ÷ 2) = 43.

Therefore we can say that 43 items are being purchased.

6 0

3 years ago

5-minute call costs $5.91 and a 10-minute call costs $10.86. what is the rate of change

KatRina [158]

Answer:

0.96 cents

Step-by-step explanation:

3 0

3 years ago

Find the common difference

Eddi Din [679]

9514 1404 393

Answer:

(a) 1/12

Step-by-step explanation:

The difference of adjacent terms is ...

1/4 -1/6 = 3/12 -2/12 = 1/12

3 0

3 years ago

Read 2 more answers

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago