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3 years ago

15

Amiya answer 19/20, of the fraction of questions correctly on her last exam. What percent of questions did Amiya answer correct

ly on her last exam?

2 answers:

larisa86 [58]3 years ago

8 0

Answer:

95%

Step-by-step explanation:

Burka [1]3 years ago

6 0

Amiya got a 95% on her test

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Suppose that an internal report submitted to the managers at a bank in Boston showed that with 95 percent confidence, the propor

Dmitry [639]

Answer:

The sample size used to compute the 95% confidence interval is 1066.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The 95% confidence interval for proportion of the bank's customers who also have accounts at one or more other banks is (0.45, 0.51).

To compute the sample size used we first need to compute the sample proportion value.

The value of sample proportion is:

$\hat p=\frac{Upper\ limit+Lower\ limit}{2}=\frac{0.45+0.51}{2}=0.48$

Now compute the value of margin of error as follows:

$MOE=\frac{Upper\ limit-Lower\ limit}{2}=\frac{0.51-0.45}{2}=0.03$

The critical value of <em>z</em> for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the value of sample size as follows:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\0.03=1.96\times \sqrt{\frac{0.48(1-0.48)}{n}}\\(\frac{0.03}{1.96})^{2}=\frac{0.48(1-0.48)}{n}\\n=1065.404\\n\approx1066$

Thus, the sample size used to compute the 95% confidence interval is 1066.

4 0

3 years ago

katovenus [111]

Answer:

x = $\frac{7+\sqrt{47}\times i }{4}$

Step-by-step explanation:

<u>To solve quadratic systems,we always substitute one variable in terms if the other and then solve the equation.</u>

x + 2y = 6 ---------------(1)

y - 5 = $(x-2)^{2}$ ---------------(2)

y = $(x-2)^{2}$ + 5 ---------------(3)

Substitute (3) in (1) ,

x + 2( $(x-2)^{2}$ + 5 ) = 6

$(a + b)^{2}$ = $a^{2} + 2ab + b^{2}$

x + 2( $x^{2} - 4x + 4$ + 5 ) = 6

$2x^{2} - 7x + 12=0$ --------------(4)

The roots of the quadratic equation $ax^{2} +bx+c$ is

x = $\frac{(-b) + \sqrt{(-b)^{2}-4 \times ac } }{2 \times a}$ -----------(5)

According to equation (5),solution of (4) is

x = $\frac{7 + \sqrt{(-7)^{2}-4 \times 24 } }{2 \times 2}$

x = $\frac{7+\sqrt{49-96}}{4}$

x = $\frac{7+\sqrt{47}\times i }{4}$

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