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3 years ago

15 - (-7) = ?? Please help if you know the answer can you maybe show work or explane how you got the answer thank you :)

Mathematics

1 answer:

Lena [83]3 years ago

5 0

Answer:

Step-by-step explanation:

since a negative times a negative equals a postive the equation becomes 15+7 which is 22

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The troposphere extends from the Earth's surface to a height of 6–12 miles, depending on the location and the season. If a plane

Anni [7]

Answer:

2.8 miles higher the plane go without leaving the troposphere?

Step-by-step explanation:

Troposphere height on different locations= 6-12 miles

plane altitude at present flight = 5.8 miles.

Troposphere deepness from the present flight= 8.6 miles

So from present altitude till troposphere= 8.6-5.8= 2.8 miles

8 0

3 years ago

Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon en

levacccp [35]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis is:

$H_o: \mu_c \ge \mu_{us}$

$H_a : \mu_c < \mu_{us}$

The t- student test statistics can be computed as:

$t = \dfrac{x_c- x_{us}}{\sqrt{\dfrac{\sigma_c^2}{n_c} + \dfrac{\sigma_{us}^2}{n_{us}} }}$

$t = \dfrac{19- 21}{\sqrt{\dfrac{7^2}{100} + \dfrac{8^2}{130} }}$

t = -2.017

degree of freedom = (n₁ - 1) + (n₂ - 1)

= (100 - 1) + (130 - 1)

= 228

Using the data of t-value and degree of freedom;

The P-value = -0.224

Decision rule: Do not reject the null hypothesis if the p-value is greater than ∝(0.01)

Conclusion: We reject the null hypothesis since the p-value is less than ∝.

Therefore, there is enough evidence to conclude that the mean age of entering prostitution in Canada is lower than that of the United States.

5 0

3 years ago

Solve 2x^2 + 26 = 0 to identify the roots

jok3333 [9.3K]

2x^2=-26
x^2=-13
x=(+/-)sqrt(13)i

4 0

3 years ago

Read 2 more answers

Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)

baherus [9]

Step 1

Given;

$4x^2-y^2+24x+4y+28=0$

Required; To find the center that eliminates the linear terms

Step 2

$\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}$

Step 3

Substitute a,d,e into the vertex form

$\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}$ $\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\ \\ \end{gathered}$

Step 4

Completing the square for -y²+4y

$\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}$ $\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}$ $\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\ \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}$

Step 5

Substitute a,d,e into the vertex form

$\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}$

Step 6

$\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}$

Step 7

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}$

Hence the answer is (-3,2)

4 0

1 year ago

Solve xy^m=yx^3 for m

aleksklad [387]

Answer:

m = 1 + 2log(x)/log(y)

Step-by-step explanation:

Taking logarithms, you have ...

log(x) +m·log(y) = log(y) +3log(x)

m·log(y) = log(y) +2·log(x) . . . . subtract log(x)

m = (log(y) +2·log(x))/log(y) . . . divide by the coefficient of m

m = 1 +2·log(x)/log(y) . . . . . . . simplify a bit*

_____

* The "simplified" form will depend on your preference. Here, I like the integer 1 brought out because most logs are irrational. The result may be very slightly more accurate if we add 1, rather than log(y)/log(y)--depending on your calculator.

6 0

3 years ago