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REY [17]
3 years ago
8

What does S in Math means?

Mathematics
1 answer:
12345 [234]3 years ago
7 0

Step-by-step explanation:

subtract

i think this

You might be interested in
A recent study focused on the number of times men and women send a Twitter message in a day. The information is summarized below
jok3333 [9.3K]

Answer:

The value of the test statistic = 2.58

Test statistic Z = - 4.805

|Z| = 4.805 > 2.58

Null hypothesis is rejected The value of the test statistic = 2.58

There is  significant difference between in the mean number of times men and women send a Twitter message in a day

Step-by-step explanation:

Step(i):-

Sample size of men  n₁ = 25

mean of the first sample x₁⁻ = 20

Standard deviation of the first sample σ₁ = 5

Sample size of women n₂ = 30

mean of the second sample x₂⁻ = 30

Standard deviation of the first sample σ₂ = 10

Level of significance ∝= 0.01

<u>Step(ii)</u>:-

Null Hypothesis : H₀: There is no significant difference between in the mean number of times men and women send a Twitter message in a day

Alternative Hypothesis :H₁:There is  significant difference between in the mean number of times men and women send a Twitter message in a day

Test statistic

Z = \frac{x^{-} _{1} - x^{-} _{2} }{\sqrt{\frac{S.D_{1} ^{2} }{n_{1} }+\frac{ S.D_{2} ^{2}}{n_{2} }  } }

Z = \frac{20 - 30 }{\sqrt{\frac{(5)^{2}  }{25 }+\frac{ (10)^{2}  }{ 30}  } }

Z =  \frac{-10}{2.081} = - 4.805

The value of the test statistic = 2.58 C

|Z| = 4.805 > 2.58

Null hypothesis is rejected The value of the test statistic = 2.58

<u>Conclusion:</u>-

There is  significant difference between in the mean number of times men and women send a Twitter message in a day

7 0
3 years ago
If a rectangle A has sides are three times the length of those rectangle B, how do the areas of the two rectangles compare?
Radda [10]

Answer:

The area of A is nine times the area of B.

Step-by-step explanation:

4 0
3 years ago
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l
Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal(\mu=4.5,\sigma^{2} =0.70^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 4.5 minutes

           \sigma = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X \leq 4.50 min)

    P(X < 5.30 min) = P( \frac{X-\mu}{\sigma} < \frac{5.30-4.5}{0.7} ) = P(Z < 1.14) = 0.8729

    P(X \leq 4.50 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.5-4.5}{0.7} ) = P(Z \leq 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( \frac{X-\mu}{\sigma} > \frac{5.30-4.5}{0.7} ) = P(Z > 1.14) = 1 - P(Z \leq 1.14)

                                                              = 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 5.30 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 5.30 min) = P( \frac{X-\mu}{\sigma} \leq \frac{5.30-4.5}{0.7} ) = P(Z \leq 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 4.00 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 4.00 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.0-4.5}{0.7} ) = P(Z \leq -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( \frac{X-\mu}{\sigma} > \frac{x-4.5}{0.7} ) = 0.05

            P(Z > \frac{x-4.5}{0.7} ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      \frac{x-4.5}{0.7}=1.645

                      {x-4.5}{}=1.645 \times 0.7

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0
3 years ago
In the distribution, the first quartile,median and mean are 30.8,48.5 and 42.0 respectively. If the co efficient skewness is -0.
finlep [7]

Answer:

The third quartile is 56.45

Step-by-step explanation:

The given parameters are;

The first quartile, Q₁ = 30.8

The median or second quartile, Q₂ = 48.5

The mean, \bar x = 42.0

Coefficient of skewness = -0.38

The Bowley's coefficient of skewness (SK) is given as follows;

SK = \dfrac{Q_3 + Q_1 - 2 \times Q_2}{Q_3 - Q_1}

Plugging in the values, we have;

-0.38 = \dfrac{Q_3 + 30.8 - 2 \times 48.5}{Q_3 - 30.8}

Which gives;

-0.38×(Q₃ - 30.8) = Q₃ + 30.8 - 2 × 48.5

11.704 - 0.38·Q₃ = Q₃ - 66.2

1.38·Q₃ = 11.704 + 66.2 =  77.904

Q₃ = 56.45

The third quartile = 56.45.

5 0
3 years ago
1. If mPCV = 42, then mZVCN =
zalisa [80]

Answer:

Easy lol mzvcn=48

Step-by-step explanation:

7 0
2 years ago
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