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gulaghasi [49]
3 years ago
13

Help meeeeeeeeeeeeeeeeeeeeeeeee

Mathematics
2 answers:
Sauron [17]3 years ago
7 0

Answer:

118.75

Step-by-step explanation:

hope it helps

...

IRINA_888 [86]3 years ago
3 0

Answer:

118.75

Step-by-step explanation:

47.50 multiplied by 2.5 would equal 118.75

ur welcome :)

You might be interested in
4 questions 20 points
Dvinal [7]
1. Cross multiply
35x = 5(11)
35x = 55
Divide both sides by 35
x = 55/35
x = 11/7

2.  (x - 2)/x = 3/8
Cross multiply
3x = 8(x - 2)
3x = 8x - 16
Subtract 8x from both sides
-5x = -16
divide both sides by -5
x = -16/-5
x = 16/5 OR  3 1/5

3.  (a + 1)/(a - 1) = 5/6
cross multiply
6(a + 1) = 5(a - 1)
distribute
6a + 6 = 5a - 5
subtract 5a from both sides
a + 6 = -5
subtract 6 from both sides
a = -11

4.  (1/3)x - 4 = (2/3)x + 6
multiply each term by 3 to clear the fractions
x - 12 = 2x + 18
subtract x from both sides
-12 = x + 18
subtract 18 from both sides
-30 = x


3 0
4 years ago
8|7+(-10)<br> Simplify:<br> 2-32
amid [387]

Answer:

8|7+(-10)=24  2-32= − 30

Step-by-step explanation:

6 0
3 years ago
today's local newspaper lists 20 stocks of local interest. of those stocks ten increased five decreased and five remained unchan
coldgirl [10]

Answer:

P(x=2) = 0.2368

Step-by-step explanation:

given data

stocks of local interest N = 20

stocks increased K = 10

stocks decreased =  5

remained unchanged = 5

solution

we take here no of stock that are increased randomly n = 2

so here Hypergeometric random variable can take integer value

{max(0, n+k - N ), min(n,K) } = {0,2}

so here

P(x=2)

so

P(x=k)  = \frac{(^K_k)\times (^{N-K}_{n-k}) }{(^N_n)}      .......................1

P(x=2 ) = \frac{(^{10}_2)\times (^{20-10}_{2-2}) }{(^{20}_2)}    

solve it we get

P(x=2 ) = 0.236842

now we use excel function as

HYPGEOM.DIST (k,n,N.cumulative)   .....................2

so it will be

HYPGEOM.DIST (2,2,10,,false)

we get

HYPGEOM.DIST (2,2,10,,false)  = 0.236842

so

P(x=2) = 0.2368

4 0
3 years ago
A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for
tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of <em>t</em> are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29

s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

  =\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}

  =-4.123

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}

   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

   =26.55\\\approx 27

Compute the critical value for <em>α</em> = 0.05 as follows:

t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

3 0
3 years ago
What is x^2 equal to? (Please don't tell me it is a graph) Is it equal to -x, 2x, or something?
Lera25 [3.4K]
x^{2} is equal to 2x, since it is the derivative of x^{2}.
3 0
3 years ago
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