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3 years ago

HELP IM LITERALLY SOBBING

Mathematics

2 answers:

Elena L [17]3 years ago

7 0

Answer:

y=1/4x+2.75

Step-by-step explanation:

two coordinates are: (5,4) and (-3,2)

the linear function is y=mx+b

m=slope

b=y-intercept

first, find the slope

the formula to find the slope is y2-y1/x2-x1

2-4/-3-5=

-2/-8=

1/4

Now you know 1/4 is the slope. Plug that into the function

y=1/4x+b

Now find the y-intercept

to do that, take one of your coordinates and plug the x and y values into the function.

I will use (-3,2)

2=1/4(-3)+b

2=-3/4+b

11/4=b

(or 2.75 decimal form)

Plug the y-intercept into the function and you're done

y=1/4x+2.75

Vika [28.1K]3 years ago

3 0

Answer: Y= 8x + 2.8

Step-by-step explanation: slope is 8 (think rise over run) and y-intercept seems to be 2.8, heh hope that helped

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2.98 times 10 to the 3rd power in standard notation

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Answer:

2980

Step-by-step explanation:

2.98*10^3 = 2980

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4 years ago

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Which of the following equations represents a line that is parallel to the line with the equation y = 2/3x +2

Andreyy89

Answer: C. 4x - 6y = 3

Step-by-step explanation:

We only need to find a line with the same slope.

We will put 4x - 6y = 3 in slope intercept

4x - 6y = 3

-6y = -4x + 3

y = 4/6 - 3/6

y = 2/3 - 1/2

There you have it, a line with the same slope, which means it's parallel.

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2 years ago

Which data set could be represented by the box plot shown below?

guapka [62]

I believe the answer is A because 32 is the median, shown by the middle line

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3 years ago

The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The

Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let X = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

(a)

The general form of a (1 - α)% confidence interval is:

$CI=SS\pm CV\times SD$

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a z-confidence interval.

The critical value of z for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the confidence interval as follows:

$CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)$

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

$37\pm 0.1=SS\pm CV\times SD$

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

$P(-1$

The percentage of z-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the z-interval or t-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a t-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For n = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - α)% t-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The critical value of t is:

$t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262$

*Use a t-table for the critical value.

The 95% confidence interval is:

$CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)$

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

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3 years ago

Which are the roots of the quadratic function f(q) = q^2 - 125? Select two options.

liubo4ka [24]

Answer:

None of the options are correct

Step-by-step explanation:

Given

$f(q) = q^2 - 125$

Required

The roots of the function

Since the function is a quadratic function; to get the roots of the function, f(q) must be equal to 0

$f(q) = q^2 - 125$ becomes

$0 = q^2 - 125$

Make $q^2$ the subject of formula

$125= q^2$

Rearrange

$q^2 = 125$

Take square roots of both sides

$\sqrt{q^2} = \sqrt{125}$

$q = \sqrt{125}$

Expand the square root of 125

$q = \sqrt{25 * 5}$

$q = \sqrt{25} * \sqrt{5}$

q = ±5 $* \sqrt{5}$

Split into 2

$q = 5 * \sqrt{5}$ or $q = -5 * \sqrt{5}$

$q = 5 \sqrt{5}$ or $q = -5 \sqrt{5}$

Hence, the roots of the quadratic function are $q = 5 \sqrt{5}$ or $q = -5 \sqrt{5}$

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3 years ago

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