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tekilochka [14]
3 years ago
6

JAYDEN MADE SOME CANDLES THAT EACH WEIGHS 1/2 POUND. HE SHIP THEM IN A BOX THAT WEIGHT 6 POUNDS. THE TOTAL WEIGHT OF THE BOX FIL

LED WITH CANDLES WAS 14 POUNDS, HOW MANY CANDLES DID JAYDEN SHIP IN THE BOX?
Mathematics
1 answer:
Ahat [919]3 years ago
6 0

Answer: 12

Step-by-step explanation:

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According to government data, 75% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women
blagie [28]

Answer:

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

We will use binomial

Formula : P(X=r) =^nC_r p^r q^{n-r}

At x = 2

P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}

P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}

P(X=2) =8.8009 \times 10^{-7}

b. That at most 2 of them have never been married?

At most two means at x = 0 ,1 , 2

So,  P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}

 P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})

P(X=r) =9.9439 \times 10^{-6}

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})

=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})

=0.2360

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3 years ago
Determinar el área de un hexágono que esta inscrito en una circunferencia de 4 unidades de radio
Dmitrij [34]

Answer:

espero y te sirva hay esta todo lo que necesitas

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2 years ago
Can you help me simplify the expression 1.452/0.6
Anastasy [175]

Answer:

1.452/0.6 simplified is 1.000/1.0

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In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.
bezimeni [28]

Given:

In the circle, m(\overarc{VUX})=152^\circ and m(\angle MUV)=77^\circ.

To find:

The following measures:

(a) m\angle VUX

(b) m\angle UXW

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

m(VUX)=2\times m\angle VWX

152^\circ=2\times m\angle VWX

\dfrac{152^\circ}{2}=m\angle VWX

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In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

m\angle VUX+m\angle VWX=180^\circ          [Opposite angles of a cyclic quadrilateral]

m\angle VUX+76^\circ=180^\circ

m\angle VUX=180^\circ-76^\circ

m\angle VUX=104^\circ

Now,

m\angle UXW+m\angle UVW=180^\circ          [Opposite angles of a cyclic quadrilateral]

m\angle UXW+77^\circ =180^\circ

m\angle UXW=180^\circ-77^\circ

m\angle UXW=103^\circ

Therefore, m\angle VUX=104^\circ  and m\angle UXW=103^\circ .

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3 years ago
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