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2 years ago

Keisha just bought 8 bags of 16 cookies each. She already had 6 cookies in a jar. How many cookies does Keisha have now?

Mathematics

2 answers:

Papessa [141]2 years ago

7 0

Answer:

134 Cookies

Step-by-step explanation:

you first multiply 8 with 16

then you add six to the product

bija089 [108]2 years ago

3 0

Answer:

134

Step-by-step explanation:

She bought 8 bags with 16 each .

So we get 8x16 , which gives us , 128 .

She already had 6 cookies so we get our total from the ones she bought and add that to out initial value , 6 .

128+6= 134

In Conclusion Keisha has 134 cookies in all .

- I really hope this helped , I'm terribly sorry if it didn't. :,)

Stay safe and have a wonderful day ! <3

~Sleepy~

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Suppose you have a bag containing 2 black marbles and 3 red marbles. You reach into the bag, select a marble, see what color it

TiliK225 [7]

Answer:

$\dfrac{9}{25}$

Step-by-step explanation:

Given that the bag contains black and red marbles.

Number of black marbles in the bag = 2

Number of red marbles in the bag = 3

Total number of marbles in the bag = Number of black marbles + Number of red marbles = 2 + 3 = 5

Let us have a look at the formula for probability of an event E, which can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(\text{First red marble}) = \dfrac{\text{Number of red marbles}}{\text{Total number of marbles}} = \dfrac{3}{5 }$

Now, the marble chosen at first is replaced.

Therefore, the count remains the same.

$P(\text{Second red marble}) = \dfrac{\text{Number of red marbles}}{\text{Total number of marbles}} = \dfrac{3}{5}$

Now, the <em>required probability</em> can be found as:

$P(\text{First red marble})\times P(\text{Second red marble}) = \dfrac{3}{5}\times \dfrac{3}{5} = \bold{\dfrac{9}{25} }$

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Step-by-step explanation:

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Step-by-step explanation:

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Half of a set of the parts are manufactured by machine A and half by machine B. Ten percent of all the parts are defective. Six

baherus [9]

Answer:

The probability that a part was manufactured on machine A is 0.3

Step-by-step explanation:

Consider the provided information.

It is given that Half of a set of parts are manufactured by machine A and half by machine B.

P(A)=0.5

Let d represents the probability that part is defective.

Ten percent of all the parts are defective.

P(d) = 0.10

Six percent of the parts manufactured on machine A are defective.

P(d|A)=0.06

Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective :

$P(A|d) =\frac{P(A \cap d)}{P(d)}$

$P(A|d) =\frac{P(d|A)\times P(A)}{P(d)}\\P(A|d)= \frac{0.06\times 0.5}{0.10}$

$P(A|d)= 0.3$

Hence, the probability that a part was manufactured on machine A is 0.3

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