The correct way to read this number 75.09

Sergey is solving 5x2 + 20% – 7 = 0. Which steps could he use to solve the quadratic equation by completing the

Mrac [35]

Answer:

$5(x^2+4x)=7$

$5(x^2+4x+4)=7+20$

$(x+2)=\pm\sqrt{\frac{27}{5}}$

Step-by-step explanation:

we have

$5x^2+20x-7=0$

step 1

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$5x^2+20x=7$

step 2

Factor 5 left side

$5(x^2+4x)=7$

step 3

Complete the square

$5(x^2+4x+2^2)=7+2^2(5)$

$5(x^2+4x+4)=7+20$

$5(x^2+4x+4)=27$

step 4

Rewrite as perfect squares

$5(x+2)^2=27$

step 5

$(x+2)^2=\frac{27}{5}$

$(x+2)=\pm\sqrt{\frac{27}{5}}$

$(x+2)=\pm\frac{3\sqrt{15}}{5}$

step 6

$x=-2\pm\frac{3\sqrt{15}}{5}$

6 0

3 years ago

Read 2 more answers

Which steps transform the graph of y = x2 to y = –2(x – 2)2 + 2? (1 point)

Kitty [74]

Y = x² to: y = - 2 ( x - 2 )² + 2
Answer:
D ) reflect across the x-axis, translate 2 units to the right, translate up 2 units, stretch by the factor of 2

7 0

2 years ago

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.