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Olenka [21]
3 years ago
8

-s/3 is equal to or greater than 6 Whoever answers get brainliest answer

Mathematics
2 answers:
kow [346]3 years ago
7 0
-s/3 ≥ 6
If you wanted the work.
Hope this helps!
sukhopar [10]3 years ago
4 0
- \dfrac{s}{3} \geq 6

Multiply both sides  by 3:
-s \geq 18

Divde both sides by -1:
x \leq 18

<em>Note: If you divide </em><em>an equality</em><em> with a negative number, the sign gets flipped.</em>
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Which equation is not equal to 6^3/6^6 ?: 1/6^2 6^-3 1/216 1/6^3
SCORPION-xisa [38]
The answer would be 60 because its multiplication all those numbers

4 0
3 years ago
14×12 distributive property​
miss Akunina [59]
186 if that’s the answer you where looking for if not comment here it will give me a notification and I will see
5 0
4 years ago
Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame
lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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 = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

Z = \frac{207.9 - 208}{0.1678}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

6 0
3 years ago
Which expression is equivalent to this expression?
saw5 [17]

Answer:

3h-4.5

Step-by-step explanation:

3/4(4h-6)

12/4h-18/4

3h-4.5

hope it's helpful ❤❤❤❤❤❤

THANK YOU.

#

4 0
3 years ago
Charles will babysit for up to 4 hours and charges $7 per hour. Write a function in function notation for this situation.
I am Lyosha [343]
Assume Charles charges $7*3=$21 if he babysits for 2,5 hours (that is 2 and a half hours).

That is we are not considering Charles babysitting exactly 1, 2, 3 or 4 hours, but also in between times.


then we can write the following piecewise function to describe the situation:


f is a function from (0, 4] to {$7, $14, $21, $28}

f(x)=\begin{cases} &#10;      \$7 & if \enspace \enspace x\leq 1 \\&#10;      \$14 & if \enspace \enspace 1\ \textless \  x\leq 2 \\&#10;      \$21 & if \enspace\enspace 2\ \textless \ x\leq 3 \\&#10;      \$28 & if \enspace\enspace 3\ \textless \ x\leq 4 \&#10;   \end{cases}
6 0
3 years ago
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