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Liula [17]
3 years ago
15

Abdallah earns AED 30 for each car he washes. He also gives his little brother AED 20 form his weekly earnings. This week, he wa

nts to have at least / minimum AED500 in spending money. Which inequality represents the number of cars that he must wash?Single choice.
A.30 x + 20 > 500

B.-30x+20 <500

C.30x -20 ≥ 500

D.20x +30≥500
Mathematics
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

Step-by-step explanation:

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An independent-measures research study uses two samples, each with n = 12 participants.If the data produce a t statistic of t =
lana [24]

Answer:

B- It cannot be answered without additional information.

Step-by-step explanation:

From the question, we have α= 0.01 and α= 0.05.

Thus;

We can decide to look at the critical values and check if test statistic is less or more than the tabulated value for α= 0.01 and 0.05.

However, that will not be sufficient as neither the level of significance nor confidence level is clearly mentioned in the question and thus, we can not answer it without additional information.

7 0
3 years ago
Angelique draws a triangle GHK. If
kap26 [50]
Where is the rest of the problem?



5 0
3 years ago
Consider the following.f(x)=74 – x2Find the x-values at which f is not continuous. Which of the discontinuities are removable?
tensa zangetsu [6.8K]

Answer

given,

f(x) = 74 - x²

A function is continuous when it is differential at all the points.

f'(x) =\dfrac{d}{dx}(74-x^2)

f'(x) =- 2 x

so, the given function is continuous for all the value of x.

We should know that if a function is quadratic then it is differential at all the point.  And if it is differential then it is definitely continuous.

Hence, there is no discontinuous point present in the equation.

6 0
3 years ago
Find the minimum and maximum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f
nata0808 [166]
Via Lagrange multipliers:

L(x,y,\lambd)=x^2y+x+y+\lambda(xy-5)
L_x=2xy+1+\lambda y=0
L_y=x^2+1+\lambda x=0
L_\lambda=xy-5=0

\underbrace{10}_{2xy}+1+\lambda y=0\implies \lambda=-\dfrac{11}y
xy=5\implies y=\dfrac5x\implies\lambda=-\dfrac{11}5x

\impliesx^2+1+\left(-\dfrac{11}5x\right)x=0\implies x^2=\dfrac56\implies x=\pm\sqrt{\dfrac56}
xy=5\implies y=\pm\sqrt{30}

At these points, we get local minima of f\left(\pm\sqrt{\dfrac56},\pm\sqrt{30}\right)=\pm2\sqrt{30}.

- - -

Another way to do this is to make f(x,y) a function independent of y, which is made possible by the constraint.

xy=5\implies y=\dfrac5x
\implies f(x,y)=f\left(x,\dfrac5x\right)=F(x)=6x+\dfrac5x
\implies F'(x)=6-\dfrac5{x^2}=0\implies x=\pm\sqrt{\dfrac56}

and so on.
8 0
3 years ago
There are 12 contestants in a coloring contest. Three ribbons will be awarded. How many different ways could the ribbons be awar
Elanso [62]

Answer:

1320 ways

Step-by-step explanation:

Here we have a situation where 3 ribbons will be awarded to 3 of the 12 contestants. We can use the permutation formula, because the order of awarding the ribbons matters.

The permutation formula is:

_{n}P_{k}=\dfrac{n!}{(n-k)!}

n = 12

k = 3

Now that we have our variables, let's plug them into the formula.

_{12}P_{3}=\dfrac{12!}{(12-3)!}

_{12}P_{3}=\dfrac{12!}{9!}

_{12}P_{3}=1320

So there are 1320 different ways that the contestants will be awarded.

5 0
3 years ago
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