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3 years ago

(-7x²-7x-7)-[(7x³-9x²-20)+(4x-11)]]=?

1 answer:

7 0

$(-7x^2-7x-7)-[(7x^3-9x^2-20)+(4x-11)]]=\\=-7x^2-7x-7-(7x^3-9x^2-20+4x-11)=\\=-7x^2-7x-7-7x^3+9x^2+20-4x+11=-7x^3+2x^2-11x+24\\\\
(5x^4+8x^2+2)+7(5x^4-3x^2)=5x^4+8x^2+2+35x^4-21x^2=\\=39x^4-13x^2+2$

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Write the multipulcation equation you can also use to find 1/3 of 1/5

ehidna [41]

Answer:

0.33 × 0.2 = 0.066 or $\frac{1}{3}$ × $\frac{1}{5}$ = $\frac{1}{15}$

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2 years ago

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Which shows the correct first step to solving the system of equations in the most efficient manner? 3 x + 2 y = 17. x + 4 y = 19

Anton [14]

Answer:

a.)x = negative 4 y + 19

Step-by-step explanation:

i got it right on e2020 lol

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4 years ago

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A) Model this situation with a linear system:

sergejj [24]

Answer:

The daily rate is $ 29 and the fee for each kilometer driven is $ 0.35.

Step-by-step explanation:

Since to rent a car, a person is charged a daily rate and a fee for each kilometer driven, and when Chena rented a car for 6 days and drove 320 km, the charge was $ 286.00, while when she rented the same car for 10 days and drove 900 km, the charge was $ 605.00, to determine the daily rate and the fee for each kilometer driven the following calculation must be performed:

286 = 6 days and 320 km

320/6 = 53,333 km

286/6 = $ 47,666

Thus, the value of 1 day and 53,333 km is $ 47,666.

605 = 10 days and 900 km

900/10 = 90 km

605/10 = $ 60.5

Thus, the value of 1 day and 90 km is $ 60.5.

90 - 53,333 = 36,666

60.5 - 47.666 = 12.833

12,833 / 36,666 = value per kilometer traveled

0.35 = value per kilometer traveled

(286 - (0.35 x 320)) / 6 = X

(286/112) / 6 = X

174/6 = X

29 = X

(605 - (0.35 x 900)) / 10 = X

(605 - 315) / 10 = X

290/10 = X

29 = X

Therefore, the daily rate is $ 29 and the fee for each kilometer driven is $ 0.35.

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3 years ago

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries

Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

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

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = $\frac{\sum X}{n}$ = 26 hr

s = sample standard deviation = $\sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} }$ = 1.625 hr

n = sample of batteries = 10

$\mu$ = population mean battery life

Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, $\mu$ is;

P(-1.833 < $t_9$ ) = 0.95 {As the lower critical value of t at 9