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3 years ago

PLEASEEEEEEEEEEEEE HELPPPPPPPPPPPPPPP AM GOING TO FAIL

Mathematics

1 answer:

sdas [7]3 years ago

7 0

Answer:

What're the answers provided in the drop-down menu?

Step-by-step explanation:

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Write the first four terms of sequence Rule:start at 28.6, subtract 3.1

Gekata [30.6K]

28.6, 25.5, 22.4, 19.3... hope this helps!!

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3 years ago

Determine the point estimate of the population proportion, the margin of error for the following confidence interval, and the

PilotLPTM [1.2K]

Answer:

point of estimate p = x/n

Step-by-step explanation:

p is the point estimate for the population , x is the number of success, n is the sample size.

p = Upper bound + Lower bond/ 2 for population proportion

x = Upper bond + Lower bond /2 for population mean

E = Upper bond - Lower bond /2 E is margin of error

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3 years ago

Scott found a rare comic book on an online auction site priced at 35 Australian dollars. If the exchange rate is $1 U.S=$1.09 Au

adell [148]

1 US = 1.09 Australian

divide by 1.09 on both sides to find how much a 1 Australian dollar is worth.

1/1.09 = 1 Australian dollar
1 Australian dollar = 0.92 US dollar

mulitply 35 on both sides of the equation

35 Australian = 35 (0.92) = 32.2 US dollars

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3 years ago

Whats the GCF of 15 and 35 and 70

Andrej [43]

Just think what do all the numbers have in common? What number can relate to all of them and is the largest number? That's how you find GCF, straight of the bat I can see that the GCF is 5 because nothing higher than five can match up to all of them, in other words nothing other than 5 or higher than 5 is a factor of all the numbers.

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3 years ago

Please check for me and if any are incorrect please explain!

Anika [276]

Answer: (1) 14.4% (2) 3.78% (3) $222.48 (4) 21,176.47

(5) $425 (6) 14.3 cents per mile

<u>Step-by-step explanation:</u>

1) This is a calculator question. I = 0.144 --> 14.4%

$A=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\bullet \text{A = Accrued amount (total amount paid)}\\\bullet \text{P = Principal (initial cost of the car)}\\\bullet \text{r = rate (interest rate in decimal form)}\\\bullet \text{n = number of times in a year (number of months)}\\\bullet \text{t = number of years}\\\\A = m \times nt\\\bullet \text{m = monthly payment amount}\\\bullet \text{nt = number of payments made}\\\implies m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}$

2) m = 93.33

nt = 36

P = 3000

n = 12

$m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\\93.33\times 36=3000\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\3359.88=3000\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\1.11996=\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\1.00315198=1+\dfrac{r}{12}\\\\\\0.00315198=\dfrac{r}{12}\\\\\\0.0378=r\\\\\\\large\boxed{3.78\%}=r$

3) <em>same equation as #2 but deduct the down payment from the Principal</em>

nt = 42

P - d = 5,555 - 555 = 5,000

r = 18% --> 0.18

n = 12

$m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\\m\times 42=(5,555-555)\bigg(1+\dfrac{.18}{12}\bigg)^{42}\\\\\\42m=5000(1.015)^{42}\\\\\\42m=5000(1.868847)\\\\\\42m=9344.23557\\\\\\m=\large\boxed{222.48}$

4) Gas + Oil = cost per mile × number of miles

1000 + 800 = 0.085x

1800 = 0.085x

$\large\boxed{21,176.47}=x$

5) $\text{Annual Depreciation}=\dfrac{\text{Cost of car - Trade-in value}}{\text{Years driven}}$

$.\qquad =\dfrac{3800-1250}{6}\\\\\\.\qquad =\dfrac{2550}{6}\\\\\\.\qquad = \large\boxed{425}$

6) $\dfrac{\text{(Cost of car - Market value) + Gas & Oil + Parts, Maintenance + Insurance}}{\text{Miles driven}}$

$=\dfrac{(5000-3800)+750+250+300}{17,500}\\\\\\=\dfrac{2500}{17,500}\\\\\\=0.142857\\\\=\large\boxed{14.3\ cents\ per\ mile}$

4 0

3 years ago