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rusak2
10 days ago
13

Harry owes the bank money. To repay his debt, he paid \$150$150dollar sign, 150 back to the bank each month. After 101010 months

, his remaining debt was \$6900$6900dollar sign, 6900.
Mathematics
1 answer:
Zina [11.9K]10 days ago
7 0


$8,400
Explanation:
Monthly repayment = $150
Total Months of repayment = 10 months
Remaining balance after 10 months = $6900
Cumulative payment = $150 × 10 = $1500
Balance remaining = $6900
Overall debt amount:
(Remaining amount + total repaid)
$(6900 + 1500)
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Micah rows his boat on a river 4.48 miles downstream, with the current , 0.32 hours. He rows back upstream the same distance, ag
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Short Answer: Current speed = 3 miles per hour. Given details for downstream distance of 4.48 miles at time 0.32 hours and upstream distance of the same 4.48 miles taking 0.56 hours. Using the equation d = r*t, we equate distances for both directions leading to a function in terms of the current speed. With each correction to solve ultimately yields the current speed as 3 mph.
5 0
16 days ago
A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit
Inessa [12141]

Answer: D) \$50\pm\$11.08

Step-by-step explanation:

Based on the provided information, we have

Sample size: n= 15

sample mean: \overline{x}=\$50.50

Sample standard deviation: s= $20

Since the population standard deviation is not known, we utilize a t-test.

For a significance level of 95% confidence: \alpha=1-0.95=0.05

Critical t-value : t_{n-1, \alpha/2}=t_{014,0.025}=2.145  [Using the Student's t-value table]

The required 95% confidence interval yields:-

\overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}\\\\ =\$50.50\pm(2.145)\dfrac{\$20}{\sqrt{15}}\\\\\approx \$50\pm\$11.08

Thus, the sought-after 95% confidence interval for the mean amount spent by credit card customers during their initial visit to the new store in the mall, assuming normal distribution of the spending amounts, is:

\$50\pm\$11.08

5 0
16 days ago
A low-strength children’s/adult chewable aspirin tablet contains 81 mg of aspirin per tablet. How many tablets may be prepared f
Svet_ta [12276]

Answer:

From 1 kg of aspirin, a total of 12,345 tablets can be produced.

Step-by-step explanation:

This query mentions that low-strength chewable aspirin tablets for children and adults contain 81 mg of aspirin each, asking how many tablets can be made from 1 kg of aspirin.

Considering the weight of a tablet measured in kg, we start by converting 81 mg to kg.

Each kg equals 1,000,000 mg. Hence

1 kg - 1,000,000 mg

x kg - 81 mg.

1,000,000x = 81

x = \frac{81}{1,000,000}

Therefore, x = 0.000081 kg.

Each tablet essentially contains 0.000081 kg of aspirin. To find out how many tablets this yields from 1 kg of aspirin:

1 tablet - 0.000081 kg

x tablets - 1 kg

0.000081x = 1

x = \frac{1}{0.000081}

Thus, x = 12,345 tablets.

Therefore, a total of 12,345 tablets can be manufactured from 1 kg of aspirin.

4 0
23 days ago
Billy Jo's school is selling tickets to a Fall Festival. On the first day of ticket sales the school sold 3 adult ticket and 8 s
PIT_PIT [11918]
Y = 6 Here’s a step-by-step explanation: On the initial day of ticket sales, the school sold 3 adult tickets and 8 student tickets for a total of $72. On the following day, the school collected $152 from 7 adult tickets and 16 student tickets. What is the price of a student ticket? Day 1: 3x + 8y = 72 Day 2: 7x + 16y = 152 This leads to a system of equations. By manipulating these equations, we find: Ultimately, y = 6. If you wish to continue from this point, substituting y into the equation will give the value for x.
6 0
15 days ago
Maureen McIlvoy, owner and CEO of a mail order business for wind surfing equipment and supplies, is reviewing the order filling
lawyer [12096]

Response:

Maureen's null hypothesis is, H₀: p₁ ≥ p₂.

Detailed explanation:

Maureen McIlvoy, as the owner and CEO of a mail-order business specializing in windsurfing gear, is scrutinizing the order fulfillment processes in her warehouses. Her objective is to achieve a 100% shipment rate of orders within 24 hours. Upon examining her warehouse operations, she discovers that both the East coast and West coast warehouses have not met this goal, although the East Coast warehouse has consistently outperformed its counterpart.

To verify this finding, Maureen’s team randomly sampled 200 orders from the West Coast warehouse (population 1) and 400 from the East Coast warehouse (population 2).

Of the sampled 200 orders from the West Coast warehouse, 190 were delivered within the specified time. In contrast, 372 out of 400 orders from the East Coast warehouse were processed within 24 hours.

The hypotheses can be formulated as followed:

H₀: The proportion of timely shipments from the East Coast does not exceed that from the West Coast warehouse, thus, p₁ ≥ p₂.

Hₐ: The proportion of timely shipments from the East Coast warehouse is indeed greater than that from the West Coast warehouse, stated as p₁ < p₂.

Thus, Maureen's null hypothesis becomes, H₀: p₁ ≥ p₂.

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15 days ago
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