Plz help me with the answer

a supplier sells 2 1/4 pounds of mulch for every 1 1/3 pounds of gravel. The supplier sells 172 pounds of mulch and gravel combi

laila [671]

We first write the fractions in their decimal forms and find out the ratio.

2 1/4 is equivalent to 2.25, and 1 1/3 is equivalent to 1.33

Ratio will be 2.25 / 1.33 = 1.69 or 1.7 rounded off. Since it cannot be simplified further, then the ratio of mulch to gravel will be 1.7: 1

Quantity of mulch sold → 1.7/2.7 of 172 →0.629 ×172 = 108.2 pounds

Quantity of gravel sold will be 172 -108. 2 = 63 .8 pounds

6 0

3 years ago

On the first one I know it is not Noah so please don´t put Noah. Also make sure to explain please

Gnoma [55]

The answer to the first one is Noah. It has to be. They have an equal amount of money. They both have 7 dollars and 50 cents.

6 0

3 years ago

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 111.4-cm and a standard dev

Kipish [7]

Answer:

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 111.4, \sigma = 0.5, n = 23, s = \frac{0.5}{\sqrt{23}} = 0.1043$

Find the probability that the average length of a randomly selected bundle of steel rods is between 111.2-cm and 111.4-cm.

This is the pvalue of Z when X = 111.4 subtracted by the pvalue of Z when X = 111.2. So

X = 111.4

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

By the Central Limit Theorem

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

$Z = \frac{111.4 - 111.4}{0.1043}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 111.2

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

$Z = \frac{111.2 - 111.4}{0.1043}$

$Z = -1.92$

$Z = -1.92$ has a pvalue of 0.0274.

0.5 - 0.0274 = 0.4726

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

5 0

3 years ago

What does a positive correlation tell you about the graph that compares advertising costs and sales. Would the owners of the com

MArishka [77]

A positive correlation means sales are rising because of advertising
the steeper the chart the more money is being made, so the owner would be happier with a steeper slope

5 0

3 years ago

Read 2 more answers

How do I simplify the answer of (y+3)(y-6)?

vlada-n [284]

This is actually a "product," not an "answer."

Nothing to simplify here. Instead, you are to multiply these two "binomials" together. Perhaps you've heard of the "FOIL method." If so, use it here.

y^2 + 3y - 6y - 18. Combining the middle terms, we get y^2 - 3y - 18. This is the simplest form of the product of <span>(y+3)(y-6). Done.</span>

4 0

3 years ago

Read 2 more answers