Which expression is equivalent to x + x

Salama makes fruit baskets with oranges and bananas. She uses between 8 and 16 oranges and bananas in a ratio of 3 oranges to 2

Studentka2010 [4]

Answer: There are 6 oranges and 4 bananas and there are 9 oranges and 6 bananas.

In both the cases, the ratio will be 3:2.

Step-by-step explanation:

Since we have given that

Salama makes fruit baskets with oranges and bananas.

Ratio of oranges to bananas is 3:2.

And she uses between 8 and 16 oranges and bananas.

We only take those number which is multiple of 5 between 8 and 16,

i.e. 10 and 15.

1) Let the total number of oranges and bananas = 10

Then, number of oranges is given by

Number of bananas is given by

2) Let the total number of oranges and bananas = 15

Then, number of oranges is given by

Number of bananas is given by

Hence, there are 6 oranges and 4 bananas and there are 9 oranges and 6 bananas.

In both the cases, the ratio will be 3:2.

7 0

3 years ago

A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses

Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = average age of the random sample of horses with colic = 12 yrs

$\mu$ = average age of all horses seen at the veterinary clinic = 10 yrs

$\sigma$ = standard deviation of all horses coming to the veterinary clinic = 8 yrs

n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( $\bar X$ $\geq$ 12)

P( $\bar X$ $\geq$ 12) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{12-10}{\frac{8}{\sqrt{60} } }$ ) = P(Z $\geq$ 1.94) = 1 - P(Z < 1.94)