6

Rewrite the following equation as a function of x.<br><br> 43 - 1/8y + 19 = 0

tamaranim1 [39]

Answer:

y = 496

Step-by-step explanation:

62 - 1/8y = 0

8 (62 - 1/8y ) = 8×0

496 - y = 0

496 = y

y = 496

5 0

3 years ago

In Joe's fish tank, there are 5 goldfish and 9 snails. How does the ratio 9:14 describe Joe's fish tank?

Natalija [7]

The answer is C, number of snails to the total number of animals

4 0

2 years ago

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f

Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.

The formula for the interval is:

p'₁± $Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }$

$Z_{1-\alpha /2}= Z_{0.95}= 1.648$

0.37±1.648* $\sqrt{\frac{0.37*0.63}{1000} }$

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂± $Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }$

0.48±1.648* $\sqrt{\frac{0.48*0.52}{1000} }$

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: $\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015$

Seniors: $\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016$

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0

3 years ago

Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.

natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is $y=x-x^2-2$ and the solutions are defined for all real numbers.