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Lera25 [3.4K]
3 years ago
7

Write the equation of the line in either point-slope form or slope-intercept form.

Mathematics
2 answers:
Nataliya [291]3 years ago
8 0

Answer:

a/b

Step-by-step explanation:

got right on edge

DedPeter [7]3 years ago
3 0

Answer:

y = -3x + 0.25

Step-by-step explanation:

A line has the following slope-intercept equation:

y = ax + b

In which a is the slope.

Write the equation of a line that has a slope of -3 and passes through the point (1.25, -4).

Slope of -3 means that a = -3. So

y = -3x + b

Passes through the point (1.25, -4). Means that when x = 1.25, y = -4

-4 = -3*1.25 + b

b = -3*1.25 + 4

b = 0.25

So

y = -3x + 0.25

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Between 6 p.m. and 10 a.m. the temperature Rose by 5 degrees Fahrenheit between 10 a.m. and 2 p.m. the temperature Rose by 6 deg
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Lines AB and CD are parallel. Find the measures of the three angles in triangle ABF.
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Step-by-step explanation:

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2 years ago
Read 2 more answers
Each morning Rin rides 1.5 miles to school and then rides home in the afternoon. Later in the evening, she rides to the park and
brilliants [131]

Answer:

Last equation given in the list of possible answers:

5 ( 1.5 + 1.5 + x ) = 25

Step-by-step explanation:

We need to include in the total addition of miles ridden during the week:

a) 1.5 miles to the school

b) 1.5 miles from school back home

c) x miles for the evening ride

so for the miles ridden per day we have: "1.5 +1.5 + x"

Now, since per week she does 5 days like this, then we need to multiply the expression above by 5 in order to total the number of miles she rides weekly (25 miles)

5 ( 1.5 + 1.5 + x ) = 25

And we can use this equation to find the amount "x" that Rin rides in the evening.

7 0
3 years ago
A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second rand
sesenic [268]

Answer:

The 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102

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".  

\bar X_1 =80 represent the sample mean 1

\bar X_2 =75 represent the sample mean 2

n1=25 represent the sample 1 size  

n2=36 represent the sample 2 size  

\sigma_1 =5 sample standard deviation for sample 1

\sigma_2 =3 sample standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =80-75=5

Since the Confidence is 0.94 or 94%, the value of \alpha=0.06 and \alpha/2 =0.03, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that z_{\alpha/2}=1.88  

The standard error is given by the following formula:

SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}

And replacing we have:

SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118

Confidence interval

Now we have everything in order to replace into formula (1):  

5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898  

5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102  

So on this case the 94% confidence interval would be given by 2.898 \leq \mu_1 -\mu_2 \leq 7.102  

7 0
3 years ago
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