A(n)___ is a letter used as a placeholder for an unknown value.

Mathematics

2 answers:

scoundrel [369]3 years ago

5 0

Variable
variables are any letters you use, such as in 3x, x is your variable. It holds an unknown value

GalinKa [24]3 years ago

3 0

The answer is a variable! An example of a variable is:
5x +5 =35
X= 6

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(2.3×10^3)+(4.0×10^5)<br>help

DiKsa [7]

4230000 is the answer make me it on this one also

3 0

3 years ago

Gail bought 1180 buttons to put on the shirts she she makes. She uses 6 buttons for each shirt. How shirts can Gail make with th

stepladder [879]

To answer this question, we need o divide.

Since she has 1180 buttons, we can divide that number by the amount of buttons it takes to make a shirt, 6.

1180/6=196

Hope this helps!

8 0

3 years ago

Read 2 more answers

A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40

bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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 $s = \frac{\sigma}{\sqrt{n}}$ .