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crimeas [40]
3 years ago
9

Help please will give a thanks !!

Mathematics
2 answers:
Vilka [71]3 years ago
7 0

Answer:

7/9, -0.3, 11/2

Step-by-step explanation:

Fofino [41]3 years ago
7 0

Answer:

7/9, -0.3, 11/2

Step-by-step explanation:

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Hence solve the equation x^3+x^2-6x=0<br>​
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<em>x³ + x² - 6x = 0</em>

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4 years ago
Read 2 more answers
I need help with these two please :’)
tatiyna

Answer:

59)

Therefore the height of the pyramid is 14 meter.

60)

Therefore the Diameter of Cone is 17.32 cm.

Step-by-step explanation:

59 )

Given:

Shape is Rectangular Pyramid,

Length = 3 meter

Width = 8 meter

Volume of Rectangular Pyramid = 112 cubic meter

To Find:

Height:

Solution:

We Know,

\textrm{Volume of Rectangular Pyramid}=\frac{1}{3}(Length\times Width\times Height)

Substituting the given values we have

112=\frac{1}{3}(3\times 8\times Height)\\ \\\therefore Height=\frac{336}{24}\\\\\therefore Height=14\ meter\\

Therefore the height of the pyramid is 14 meter.

60 )

Thee shape is of Cone

Given:

Height= 5 cm

Volume of cone = 125 π cm³

To Find:

Diameter of a cone = d = ?

Solution:

We know,

\textrm{Volume of cone}=\frac{1}{12}\pi (Diameter) ^{2}\times Height

Substituting the given values we have

125\pi =\frac{1}{12}\pi  d^{2}\times 5\\ \\d^{2}=25\times 12\\  \\d^{2}=300\\\\d=17.32\\   \therefore Diameter = 17.32\ cm

Therefore the Diameter of Cone is 17.32 cm.

6 0
3 years ago
Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about 6.9. Use the criterion tha
zavuch27 [327]

Answer:

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

\mu = 80, \sigma = 6.9

Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean

This means that z-scores higher than 2.5 or lower than -2.5 are considered unusual.

Would it be unusual to have more than 120 successes out of 200 trials

We have to find the Z-score of X = 120.

So

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

Z = \frac{120 - 80}{6.9}

Z = 5.8

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

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