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4 years ago

Why does -2.4-(-7)have the same result as -2.4+7? Explain your reasoning.

Mathematics

2 answers:

asambeis [7]4 years ago

8 0

Answer:

-2.4-(-7) has the same result as -2+7 because the negative sign and the minus sign cancel each other out. Think of it like this. If you cross the negative sign and the minus sign together, then it makes a plus sign.

Step-by-step explanation:

I gave a basic answer. You might have to elaborate more on this.

torisob [31]4 years ago

8 0

If you have 2 negatives it equals a positive

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1 - 4 please help !!!!

dusya [7]

We can use the slope formula for each of these.

$\frac{y.2 - y.1}{x.2 - x.1} = m$

1. 5 - 1 / 4 - 2

4 / 2

The slope is 2, and it is rising because it is positive.

2. -5 - 0 / 3 - (-1)

-5 / 4

The slope is -5/4, and it is falling because it is negative.

3. 1 - 1 / -3 - 2

0 / -5

The slope is 0, and it is a horizontal line. The value of y is not changing per value of x.

4. -5 - 2 / -1 - (-1)

-7 / 0

Undefined

The slope is undefined because it fails the vertical line test.

3 0

4 years ago

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

3 years ago

a rectangle garden has a length of (y+2) and a with of (4y-1) yards. find the perimeter of the garden on terms of y.

kobusy [5.1K]

<h3>Answer:</h3>

<u>Perimeter of rectangular garden = 2(l + b)</u>

2(y + 2 + 4y - 1)
2y + 4 + 8y - 2
10y + 2.

<u>Hence, perimeter of given garden is 10y + 2</u>.

8 0

3 years ago

What MUST be true about the triangle? A) The triangle is right. B) The triangle is scalene. C) The triangle is isosceles. D) The

Sati [7]

D) but you didnt give me enough detail

8 0

3 years ago

Read 2 more answers

What is the probability of drawing a king, then an ace (with replacement)?

alexandr1967 [171]

Answer:

So there is a 2 in 13 chance to pull out either an ace or king from the deck

Step-by-step explanation:

3 0

3 years ago