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DanielleElmas [232]

3 years ago

5

Cassie is thinking of two numbers. Adding 4 times the first number and 6 times the second number gives a total of 28. Also, addi

ng 5 times the first number and 6 times the second number gives 29. What are the two numbers?

1 answer:

aleksandrvk [35]3 years ago

3 0

If you try 6x4, you get 24. When you add 4, it = 28. The same thing with 6x4=24+5=29

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A birthday celebration meal is $ 46.80

kolezko [41]

10% of 46.80 =4.68
46.80 + 4.68 = 51.48

Hope this helps :D

3 0

2 years ago

10d - 6d = 4 find what d equals

Nutka1998 [239]

Answer:

$\huge\boxed{\sf d = 1}$

Step-by-step explanation:

10d - 6d = 4

Take d common

d (10 - 6) = 4

d (4) = 4

4d = 4

Divide both sides by 4

d = 1

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3>

5 0

3 years ago

Read 2 more answers

2. What is the value of x if the angles<br> within a triangle are 2x +1, 7x-2 and 3x<br> +5?

noname [10]

Answer:

$x=14.67\\$ °

Step-by-step explanation:

The sum of all the three angles of any triangle is 180°

Putting sum of all the angles given equals to 180°

$(2x+1)+(7x-2)+(3x+5)=$ $180$ °

$2x+7x+3x+1-2+5=180$ °

$12x+4=180$ °

$x=\frac{180-4}{12}$

$x=14.67\\$ °

The value of the x is found on solving the given angles of the triangles.

8 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

Divide using long division. check your answers. 1. (x 3 – 13x 2 + 7x – 48) ÷ (x 2 + 3) 3. (x 3 + 5x 2 – 3x – 1) ÷ (x – 1) 5. (x

artcher [175]

1. We can conclude that the result of the division is $x-13$ with a remainder of $4x-9$ , or in other notation: $x-13+ \frac{4x-9}{x^2+3}$ . Check the steps in the first image.

3. We can conclude that the result of the division is $x^{2} +6x+3$ with a remainder of 2, or in other notation: $x^{2}+6x+3+ \frac{2}{x-1}$ . Check the steps in the firts image.

5. We can conclude that the result of the division is $x+1$ with a remainder of 5, or in other notation: $x+1+ \frac{5}{x-4}$ . Check the steps in the first image.

7. To check if $x+4$ is a factor of $x^{3}+3x^{2}-10x-24$ , we are going to perform long division. If the result of the long division has no remainder, $x+4$ is a factor of $x^{3}+3x^{2}-10x-24$ .
Since the result of the long division has no remainder, we can conclude that $x+4$ is a factor of $x^{3}+3x^{2}-10x-24$ . Check the steps in the second image.

9. To check if $x-3$ is a factor of $x^{3}+3x^{2}-10x-24$ , we are going to perform long division. If the result of the long division has no remainder, $x-4$ is a factor of $x^{3}+3x^{2}-10x-24$ .
Since the result of the long division has no remainder, we can conclude that $x-4$ is a factor of $x^{3}+3x^{2}-10x-24$ . Check the steps in the third image.

10. We can conclude that the result of the synthetic division is $-2x^{2}+9x+5$ . Check the steps in the fourth image.

3 0

3 years ago