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3 years ago

Identify the coefficient of 2xy3

Mathematics

2 answers:

stiv31 [10]3 years ago

6 0

2

The coefficient always comes before any unknowns, so for 2xy3, 2 is the coefficient.

Mama L [17]3 years ago

5 0

The coefficient is 2

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Which of the following is the graph of y=3 sec [ 2(x-pi/2)] + 2

asambeis [7]

Answer:

A.) The first graph

Step-by-step explanation:

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3 years ago

You go shopping and spend $264 on shirts and shorts. You purchase a total of 9 items. Each shirt costs $24 and each pair of shor

Kisachek [45]

Answer:

3 shirts, 6 shorts

Step-by-step explanation:

A total of $264 is spent during shopping on shirts and shorts

Each shirt costs $24

Each shorts cost $32

Therefore the quantity of shirts and shorts bought can be calculated as follows

24× 3= 72

32×6= 192

192+72= 264

Hence 3 shirts and 6 shorts were bought

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3 years ago

(05.05)Polygon ABCD has the following vertices:

Kaylis [27]

Answer is option b)38.5 sq.units

Given vertices are A(-4,2) ,B(3,2),C(3,-5) and D(-4,-2)

Area of polygon is defined in the image.

If you have any doubts see the image or ask in comments box.

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4 0

3 years ago

The value of y varies directly with x2, and y = 150 when x = 5. What is the value of y when x=3?

kow [346]

The value of y when x=3 when the value of y varies directly with x^2, and y = 150 when x = 5 is 54. Since the value of y varies directly with x2, then: y = k * x^2, where k is constant. When y = 150 and x = 5, the value of constant is: 150 = k * 5^2. 150 = k * 25. k = 150/25. k = 6. Thus, when x = 3, the value of y will be: y = 6 * 3^2. y = 6 * 9. y = 54.

3 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 t 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:

H₀: There is no difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ = 0.

Hₐ: There is a difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ ≠ 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 t-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 α = 0.05 as follows:

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

*Use a t-table for the values.

Thus, the critical values of t are ± 2.052.

3 0

3 years ago