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NikAS [45]
3 years ago
7

An expression to convert 50 miles per minute is?

Mathematics
1 answer:
andrew11 [14]3 years ago
7 0

Answer:

50mpm

Step-by-step explanation:

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Simplify the expression. 7c3 • –5c7
Sidana [21]

Answer:

-35c^12

Step-by-step explanation: Hope this helped.

7 0
3 years ago
The distance d of a particle moving in a straight line is given by d(t) = 2t3 + 5t – 2, where t is given in seconds and d is mea
Sergeeva-Olga [200]

Answer:

(C)6t^2+5

Step-by-step explanation:

Given the distance, d(t) of a particle moving in a straight line at any time t is:

d(t) = 2t^3 + 5t - 2, $ where t is given in seconds and d is measured in meters.

To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).

v(t)=\dfrac{d}{dt}\\\\v(t) =\dfrac{d}{dt}(2t^3 + 5t - 2) =3(2)t^{3-1}+5t^{1-1}\\\\v(t)=6t^2+5

The correct option is C.

6 0
3 years ago
HA=2-bA , solve the equation for A
yaroslaw [1]
HA + bA = 2
A(h+b) = 2
A = 2/h+b
7 0
2 years ago
Read 2 more answers
A report states that the mean yearly salary offer for students graduating with a degree in accounting is $48,734. Suppose that a
DENIUS [597]

Solution :

To claim to be tested is whether "the mean salary is higher than 48,734".

i.e. μ > 48,734

Therefore the null and the alternative hypothesis are

$H_0 : \mu = 48,734$

and $H_1 : \mu > 48,734$

Here, n = 50

$\bar x = 49,830$

s = 3600

We take , α = 0.05

The test statistics t is given by

$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}$

$t=\frac{49,830 - 48,734}{\frac{3600}{\sqrt 50}}$

t = 2.15

Now the ">" sign in the $H_1$ sign indicates that the right tailed test

Now degree of freedom, df = n - 1

                                              = 50 - 1

                                              = 49

Therefore, the p value = 0.02

The observed p value is less than α = 0.05, therefore we reject $H_0$. Hence the mean salary that the accounting graduates are offered from the university is more than the average salary of 48,734 dollar.

3 0
3 years ago
Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and
uysha [10]

Answer:

\frac{x^{2}}{36} - \frac{y^{2}}{64}=1

Step-by-step explanation:

Given an hyperbola with the following conditions:

  • Foci at (-10,0) and (10,0)
  • x-intercept +/- 6;

The following holds:

  • The center is midway between the foci, so the center must be at (h, k) = (0, 0).
  • The foci are 10 units to either side of the center, so c = 10 and c^2 = 100
  • The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and a^2 = 36.

Then, a^2+b^2=c^2\\b^2=100-36=64

Therefore, substituting a^2 = 36. and b^2=64 into the standard form

\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1

4 0
3 years ago
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