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2 years ago

A yogurt costs 55p .How many yogurts can be bought for six pounds?

1 answer:

7 0

Well if one pound of yogurt cost 55 (cents?)

Two yogurts cost $1.10
Three yogurts cost $1.65
Four yogurts cost $2.20
Five yogurts cost $2.75
And finally...
Six yogurts cost $3.30

Hope this helps!

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A rectangular prism has a length of 3 1/2 inches, a width of 3 1/2 inches, and a height of 7 inches. Danny has a storage contain

DENIUS [597]

To find the volume of the prism you multiply
length x width x height. So for the prism you multiply 3 1/2 x 3 1/2 x 7. Which gives you the answer of 85.75, or rounded it’ll give you 86.
Now to find the difference you subtract the volume of the prism from the container which will give you your final answer

6 0

2 years ago

Read 2 more answers

topjm [15]

No she is not correct

3 0

2 years ago

75% increase followed by 50% decrease is it greater than to original

IceJOKER [234]

Answer:

Set original amount = x

<u>After a 75% increase, it would become</u>

x + 75%x = x + 0.75x = x(1 + 0.75) = 1.75x

<u>After a 50% decrease, it would become</u>

1.75x - 50%(1.75x) = 1.75x - 0.5(1.75x) = 1.75x - 0.875x = 0.875x = $\frac{7}{8} x$

Because $\frac{7}{8} x$ is less than x, the new amount would be less than the original.

7 0

2 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

2 years ago

Carly eats a bowl of granola for breakfast every morning. She wonders how much granola she eats in a year. If Carly eats about o

sp2606 [1]

36 pounds of granola

7 0

3 years ago

Read 2 more answers