19. 4x²y(x + 2) - x(xy2 – 3)

1 answer:

6 0

4x^2y(x+2)-x(xy2-3)
(4x^2y)^(x)+(4x^2y)(2)-x^2y^2+3x
4x^3y+8x^2y-x^2y^2+3x
4x^3y-x^2y^2+8x^2y+3x
This was very confusing to solve so sorry if I’m wrong.Also sorry if it’s too hard to write it.

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I have no clue how to solve. Need help ASAP

agasfer [191]

Answer:269

Step-by-step explanation:

8x18=144

6x10=60

9x5=45

5x4/2=10

144+60+45+10+10=269

8 0

3 years ago

write the equation for an exponential function, in the forms y = a x b^x, whose graph passes through the coordinate points (1, 7

zysi [14]

Y = abˣ
(a ≠ 0, b ≠ 0)

(1, 7.5)
x = 1
y = 7.5
  ↓
7.5 = ab

(3, 16.875)
x = 3
y = 16.875
  ↓
16.875 = ab³

7.5 = ab
b b
7.5/b = a

  16.875 = ab³
b³ b³
16.875/b³ = a

7.5/b = 16.875/b³
7.5/b(b³) = 16.875/b³(b³)
7.5b² = 16.875
7.5 7.5
b² = 2.25
  √(b²) = √(2.25)
b = 1.5

a = 7.5/b
a = 7.5/1.5
a = 5

y = 5(1.5)ˣ

8 0

4 years ago

Read 2 more answers

he American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 45 w

Sonja [21]

Answer:

The 85% confidence interval is (0.2187, 0.2613).

Step-by-step explanation:

We have a large sample of size n = 830 Americans over 45. We have observed that 631 people don't smoke. Let p be the population proportion of Americans over 45 who smoke. An point estimate of p is $\hat{p} = 199/830 = 0.24$ and an estimate of the standard deviation of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.24)(0.76)/830}$ = 0.0148. Therefore, because we have a large sample, the 85% confidence interval for the population proportion is given by $\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = 0.24\pm z_{0.15/2}0.0148 = 0.24\pm z_{0.075}0.0148$ where $z_{0.075}$ is the 7.5th quantile of the standard normal distribution, i.e., -1.4395. Therefore the 85% confidence interval is $0.24\pm (1.4395)(0.0148)$ or equivalently (0.2187, 0.2613).