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Serjik [45]
3 years ago
14

A Girl Scout camp served half of their granola with breakfast. After dinner, they put the remaining 6{,}250 \text{ g}6,250 g6, c

omma, 250, start text, space, g, end text of granola on top of their yogurt dessert. How many kilograms of granola did the Girl Scout camp start with?
Mathematics
1 answer:
Iteru [2.4K]3 years ago
5 0

Answer:

12.5 kilograms

Step-by-step explanation:

Given:

Half of the granola served with breakfast.

Let the amount of Granola present initially = x kg

Half of it (\frac{x}{2}) served with breakfast, so remaining granola :

x-\dfrac{x}{2} = \dfrac{x}{2}

After dinner, the remaining granola 6,250 gm granola was served with yogurt dessert.

First of all, let us convert it to kilogram because the answer is required in Kilograms.

We know that:

1000 gms = 1 kilograms

1 gms = 0.001 kilograms

6250 gms  = 6250 \times 0.001 = 6.250 kilograms

Now, as per given question statement:

\dfrac{x}{2} = 6.250\ kg\\\Rightarrow x = 2 \times 6.250\\\Rightarrow \bold{x = 12.5\ kg}

So, the answer is:

Girl scout camp started with 12.5 Kilograms of granola.

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8 0
3 years ago
Redondear a 100,000 el numero 5,370,288
Softa [21]

Para redondear este número a 1,000, verifiquemos el tercer dígito del número.

Si el dígito es mayor que 5, aumentamos el siguiente número en 1 unidad.

Si el dígito es menor o igual a 5, mantenemos el siguiente número.

El tercer dígito es 3, por lo que el número redondeado es:

25,000

6 0
1 year ago
A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0
yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

P(x

Where, for example, P(0) and P(1) are calculated as:

P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098

Then, P(x, so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)

Where, for example, P(4) is calculated as:

P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001

Then, P(x\geq4 )=0.3504, so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0
3 years ago
Raul spent b dollars for lunch. Dolly spent 1/3 of the amount that Raul spent. If c represents the amount dolly spent, express c
Andru [333]
The following problem could be expresses as 

c=1/3b     or     3c=b


4 0
3 years ago
Find the value of X<br><br> A.66<br><br> B.58<br><br> C.32<br><br> D.68
Alina [70]

1) Δ ABC

m∠B + m∠C + m∠BAC = 180⁰

2) m∠DAB + m∠BAC = 180⁰ because m∠DAB and m∠BAC are supplementary angles.

3) m∠B + m∠C + m∠BAC = m∠DAB + m∠BAC

m∠B + m∠C = m∠DAB

32⁰ + x = 98⁰

x=98 - 32 = 66 ⁰

Answer: A. 66⁰.

8 0
3 years ago
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