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Oksi-84 [34.3K]
3 years ago
10

the current temperature in smalltown is 20 °F this is 6 degrees less than twice the temperature thst that it was six hour ago wh

at was the temperature is smalltown six hout ago
Mathematics
1 answer:
ira [324]3 years ago
6 0
<span>2 x - 6 = 20, where x is the temperature in degrees Farhenheit 6 hours ago. 2 x = 20 + 6, 2 x = 26, x = 26 : 2, x = 13 degrees F. We can prove it: 13 * 2 - 6 = 26 - 6 = 20 degrees F ( the current temperature ) Answer: The temperature 6 hours ago was 13 degrees F.Hope this helps. Let me know if you need additional help!</span><span />
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£150 is divided between Nick, June &amp; Sushil so that Nick gets twice as much as June, and
BARSIC [14]

Answer:

Amount June get = £45

Step-by-step explanation:

Given:

Total amount divided = £150

Amount Nick get = 2 x Amount June get

Amount June get = 3 x Amount Sushil get

Find:

Amount divided between Nick June and Sushil

Computation:

Assume;

Amount Sushil get = a

Amount June get = 3 x Amount Sushil get

Amount June get = 3a

Amount Nick get = 2 x Amount June get

Amount Nick get = 2 x 3a

Amount Nick get = 6a

Amount Sushil get + Amount June get + Amount Nick get = Total amount

So,

a + 3a + 6a = 150

10a = 150

a = 15

So,

Amount Sushil get = £15

Amount June get = 3a

Amount June get = 3 x 15

Amount June get = £45

Amount Nick get = 6a

Amount Nick get = 6 x 15

Amount Nick get = £90

4 0
3 years ago
An axiom in Euclidean geometry states that in space, there are at least points that do
oksian1 [2.3K]

An axiom in Euclidean geometry states that in space, there are <u>2</u> points that <u>lie on the same line</u>.

This is called the two-point postulate. According to Euclidean geometry, in space, there are at least two points, and through these points, there exists exactly one line. This means that there is only one single line that could pass between any two points. This is a mathematical truth. It is known as an axiom because an axiom refers to a principle that is accepted as a truth without the need for proof.

6 0
4 years ago
Read 2 more answers
In the figure below mPUR =77 and mST=74 Find mPR
eduard

Answer:

a)

<u>The angle formed by intersecting chords is half of the sum of the intercepted arcs:</u>

  • 1/2(mST + mPR) = m∠PUR
  • 1.2(74° + mPR) = 77°
  • 74° + mPR = 2*77°
  • mPR = 154° - 74°
  • mPR = 80°

b)

<u>The angle formed by two tangents is the half of the difference of the intercepted arcs:</u>

  • m∠PSR = 1/2(mPTR - mPR)
  • m∠PSR = 1/2(247° - (360° - 247°))
  • m∠PSR = 247° - 180°
  • m∠PSR = 67°
8 0
3 years ago
Are 0,2,5 closed or not under addition
kvasek [131]

Answer:

There are closed addition.

Step-by-step explanation:

Closure (mathematics) ... For example, the positive integers are closed under addition, but not under subtraction: is not a positive integer even though both 1 and 2 are positive integers. Another example is the set containing only zero, which is closed under addition, subtraction and multiplication.

So, 0,2,5 are closed addition

Please mark me brainliest.

3 0
4 years ago
Exercise 5.2. Suppose that X has moment generating function
soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Step-by-step explanation:

Given:

moment generating function  of X as:

MX(t) = \frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}

a)  Now

Mean, E(X) = M_{X}'(t=0)

Thus,

M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}

or

M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}

also,

E(X^{2})=M_{X}''(t=0)

Thus,

M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}

or

M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}

Therefore,

Mean, E(X) = M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}

or

Mean, E(X) = - 0.5

and

E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}

or

E(X^{2}) = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Therefore,

at x = 0, P(x) = \frac{1}{2}

at x= - 4 ,P(x) = \frac{1}{3}

at x = 5, P(x) = \frac{1}{6}

Thus,

E(X) =\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})

or

E(X) = - 0.5

also,E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})

E(X^{2})  = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

4 0
4 years ago
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