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Natasha_Volkova [10]

1 year ago

7

Which inequality represents the sentence below? Four and nine tenths less than the product of a number and six is no more than e

ighteen and one-eighth. 4 and StartFraction 9 over 10 EndFraction minus 6 x less-than-or-equal-to 18 and StartFraction 1 over 8 EndFraction 6 x minus 4 and StartFraction 9 over 10 EndFraction less-than-or-equal-to 18 and StartFraction 1 over 8 EndFraction 4 and StartFraction 9 over 10 EndFraction less-than 6 x + 18 and StartFraction 1 over 8 EndFraction 6 x minus 4 and StartFraction 9 over 10 EndFraction less-than 18 and StartFraction 1 over 8 EndFraction

2 answers:

RUDIKE [14]1 year ago

7 0

Answer:

The answer is C

Step-by-step explanation: Hope this helps!

Makovka662 [10]1 year ago

4 0

Answer:

the answer is c

Step-by-step explanation:

I took the test

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QUESTION THREE (30 MARKS) 3.1 The mass of a standard loaf of white bread is, by law meant to be 700g with a population standard

kiruha [24]

Using the normal distribution and the central limit theorem, it is found that there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

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Mean of 700g means that $\mu = 700$
Standard deviation of 21g means that $\sigma = 21$
Sample of 64, thus $n = 64$
<u>For the sampling distribution of the sample mean</u>, the standard deviation is of $s = \frac{21}{\sqrt{64}} = \frac{21}{8} = 2.625$

The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:

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

By the Central Limit Theorem

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

$Z = \frac{695 - 700}{2.625}$

$Z = -1.905$

$Z = -1.905$ has a p-value of 0.0284.

0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

A similar problem is given at brainly.com/question/22934264

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shepuryov [24]

10850 would be I'm favor of building the park and 4650 would not

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1 year ago

A recent study found that 51 children who watched a commercial for Walker Crisps (potato chips) featuring a long-standing sports

hichkok12 [17]

Answer:

1

The claim is that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

2

The kind of test to use is a t -test because a t -test is used to check if there is a difference between means of a population

3

$t = 3.054$

4

The p-value is $p-value = P(Z > 3.054) = 0.0011291$

5

The conclusion is

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

The test statistics is

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 51$

The first sample mean is $\mu_1 = 36$

The second sample size is $n_2 = 41$

The second sample size is $\mu_2 = 25$

The first standard deviation is $\sigma _1 = 21.4 \ g$

The second standard deviation is $\sigma _2 = 12.8 \ g$

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \mu_1 = \mu_ 2$

The alternative hypothesis is $H_a : \mu_1 > \mu_2$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }$

=> $t = \frac{ 36 - 25}{ \sqrt{ \frac{ 21.4^2}{51} + \frac{ 12.8^2}{41} } }$

=> $t = 3.054$

The p-value is mathematically represented as

$p-value = P(Z > 3.054)$

Generally from the z table

$P(Z > 3.054) = 0.0011291$

=> $p-value = P(Z > 3.054) = 0.0011291$

From the values obtained we see that $p-value < \alpha$ so the null hypothesis is rejected

Thus the conclusion is

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

5 0

2 years ago