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lakkis [162]
3 years ago
7

Find the mean absolute deviation for the set of data. Number of sibblings 2,5,8,9,7/6,3,5,1,4,

Mathematics
1 answer:
vova2212 [387]3 years ago
8 0

the mean of 2,5,8,9,7/6,3,5,1,4, is 4.2

and the mean of 2,5,8,9,76,3,5,1,4, is 5

hope this helps have an amazing day and be safe

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Which is true regarding the graphed function f(x)?
serg [7]

Answer:

f(5)=-1

Step-by-step explanation:

4 0
3 years ago
Two 5-year girls, Alyse and Jocelyn, have been training to run a 1-mile race. Alyse's 1 mile time A is approximately Normally di
tatyana61 [14]

Answer:

1.7 × 10⁻⁴

Step-by-step explanation:

The question relates to a two sample z-test for the comparison between the means of the two samples  

The null hypothesis is H₀:  μ₁ ≤ μ₂

The alternative hypothesis is Hₐ: μ₁ > μ₂

z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}

Where;

\bar {x}_1 = 13.5

\bar {x}_2 = 12

σ₁ = 2.5

σ₂ = 1.5

We set our α level at 0.05

Therefore, our critical z = ± 1.96

For n₁ = n₂ = 23, we have;

z=\dfrac{(13.5-12)-(0)}{\sqrt{\dfrac{2.5^{2} }{23}-\dfrac{1.5^{2}}{23}}} = 3.5969

We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96

Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983

Therefore;

The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.

8 0
3 years ago
Jamal earns $40 washing 5 windows. At this rate, how many windows did Jamal wash to earn $80?​
Alja [10]
10 windows : divide 40/5=8
Then divide 80/8 =10
3 0
2 years ago
Read 2 more answers
Which value for y makes the sentence true? 8 - y = 6 - 1 A. 4 B. 5 C. 1 D. 3
Oksi-84 [34.3K]

Answer:

4

Step-by-step explanation:

3 0
3 years ago
Scores on an exam follow an approximately Normal distribution with a mean of 76.4 and a standard deviation of 6.1 points. What p
klasskru [66]

Answer:

99.89% of students scored below 95 points.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 76.4, \sigma = 6.1

What percent of students scored below 95 points?

This is the pvalue of Z when X = 95. So

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

Z = \frac{95 - 76.4}{6.1}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989.

99.89% of students scored below 95 points.

5 0
3 years ago
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