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3 years ago

How to solve (-7)+10?

Mathematics

2 answers:

slavikrds [6]3 years ago

8 0

-7 + 10 = 3
=
10 + (-7)
Is the same as
10 - 8
…
Not sure if this was that helpful.. I hope it did.

olganol [36]3 years ago

7 0

Answer:The answer will be 3

Step-by-step explanation:

First you turn the addition sigh into subtraction

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Find the slope of any line parallel to the lien through the given points (3,2),(3,1)

Alborosie

Answer:

m = undefined

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Parallel lines have the same slope but different y-intercepts
An undefined line is a vertical line
Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (3, 2)

Point (3, 1)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [Slope Formula]: $\displaystyle m=\frac{1-2}{3-3}$
[Fraction] Subtract: $\displaystyle m=\frac{-1}{0}$
Simplify: m = undefined

8 0

3 years ago

What’s a line equation passing through (14,-8) and (24,-3)

Nataly_w [17]

$\bf (\stackrel{x_1}{14}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{24}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-8)}}}{\underset{run} {\underset{x_2}{24}-\underset{x_1}{14}}}\implies \cfrac{-3+8}{10}\implies \cfrac{5}{10}\implies \cfrac{1}{2}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-8)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{14}) \\\\\\ y+8=\cfrac{1}{2}x-7\implies y=\cfrac{1}{2}x-15$

4 0

3 years ago

Find the equation of the line with slope = 2 and y-intercept = -8.

elena-s [515]

Answer: d

Step-by-step explanation:

Using the linear equation y=mx+b, you are given m (the slope) and b (the y-intercept). m=2 and b=-8. Therefore, y=2x-8

6 0

3 years ago

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

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".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.