All categories

3 years ago

8 students want to share 6 candy bars equally. How much will each student get?

Mathematics

2 answers:

Crank3 years ago

3 0

Answer: 1.3

Step-by-step explanation:

8 divided by 6 will equal 1.3 so each student will get 1.3

Stella [2.4K]3 years ago

3 0

1.3; 8/6 = 1.3 so each person will get 1 and 1/3 of a candy bar.

You might be interested in

Simplify using order of operations.Show all your work. 4^3 + (`10-8) divided by 2

Igoryamba

Answer:

Step-by-step explanation:

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

------------------------------------------------------------------------

$4^3+(10-8)\div2=64+2\div2=64+1=65$

3 0

3 years ago

You may remember that the perimeter of a rectangle is P=2(W+L) where W is the width and L is the length. Suppose that the perime

Anna71 [15]

Answer:

5 feet

Step-by-step explanation:

P = 2(w + l)

P = 34

l = w + 7

34 = 2 (w + w + 7)

34 = 2w + 2w + 14

34 - 14 = 4w

20 = 4w

w = 5 feet

Since l = w + 7

l = 5 + 7

l = 12 feet

5 0

3 years ago

Sam had monkeys and lions in his zoo, He counted legs one day to discover that he could see 76 feet on the ground. He knew that

cricket20 [7]

If he counts 76 feet, that means there’s 38 animals in total, because animals have 2 feet. 76/2=38

Hope that helps..!

5 0

3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$