Which equation when graph has x intercepts at (2,0) and (4,0) and a y-intercept of (0,-16)?

1 answer:

8 0

X intercepts are when y=0, so the intercepts are the zeroes. That gives us the possible answers A and C, but only C makes the y intercept (when x=0) = -16, so C is the answer.

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Simplify the expression and show work.

nasty-shy [4]

Answer:

72/149

Step-by-step explanation:

Just to clarify, the curly braces mean the same thing as normal parenthesis.

(12 + 5^2) = (12 + 25) = 37

(-7-15)^2 = (-22)^2 = 484

6^3 = 216

(12 + 5^2) - (-7-15)^2 = 37 - 484 = 447

6^3 / ((12 + 5^2) - (-7 - 15)^2) = 216 / 447 = 72/149

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k0ka [10]

Answer:

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3 years ago

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What is 7 times as much as the sum of 1/3 + 4/5

riadik2000 [5.3K]

Lets write the problem info into an equation and solve step by step:
7(1/3 + 4/5)
the minimum common multiple of 3 and 5 is 15, so we multiply and divide the fractions by a proper number to convert them to be divided by 15 so is easier to add them:
(1/3)(5/5) = (1*5)/(3*5) = 5/15
(4/5)(3/3) = (4*3)/(5*3) = 12/15
so we substitute in the original equation:
7(1/3 + 4/5)
= 7(5/15 + 12/15)
= 7(17/15)
= (7*17)/15
= 119/15

8 0

2 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

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".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by: