What is the standard form of the equation given the slope of:

Sonja [21]

Since 42 ÷ 6 = 7, we will round 4,189 to 4,200 and 6.3 to 6 . Estimate: ... When estimating decimal quotients, it is easiest to use compatible numbers.

8 0

3 years ago

24:36:72 simplest form

marishachu [46]

Answer:

2 : 3 : 6

Step-by-step explanation:

24:36:72

Dividing through by 12

= 24/12:36/12:72/12

= 2 : 3 : 6

8 0

3 years ago

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries

Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = $\frac{\sum X}{n}$ = 26 hr

s = sample standard deviation = $\sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} }$ = 1.625 hr

n = sample of batteries = 10

$\mu$ = population mean battery life

Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, $\mu$ is;

P(-1.833 < $t_9$ ) = 0.95 {As the lower critical value of t at 9

degrees of freedom is -1.833 with P = 5%}

P(-1.833 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $-1.833 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ ) = 0.95

P( $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ ) = 0.95

95% lower confidence interval for $\mu$ = [ $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $26-1.833 \times {\frac{1.625}{\sqrt{10} } }$ ]

= [25.06 hr]

Therefore, a 95% lower confidence interval on mean battery life is 25.06 hr.

5 0

3 years ago

Write the point-slope form of the equation of the horizontal line that passes through the point (2, 1). Pease show how!

mote1985 [20]

Check the picture below.

using a second for it, since it's a horizontal line, then say hmmm we'll use the y-intercept, or (0, 1), so the equation of a line that passes through (2,1) and (0,1)

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ 1}})\quad 
% (c,d)
&({{ 0}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-1}{0-2}\implies \cfrac{0}{-2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=0(x-2)$

5 0

3 years ago

A tourist group arrived at the boat dock for a boat ride in the morning. There were 252 people in the group. Another tourist gro

seraphim [82]

Okay, so to get the answer we have to divide.

252(morning group) divided by 36 = 7 trips

288(afternoon group) divided by 36= 8 trips

7 + 8 =
Answer: 15trips

8 0

4 years ago

What is the standard form of the equation given the slope of: - 3/5 and y-intercept 5