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2 years ago

£399 is shared between Ann, Bill, Chloe and Dave.

Mathematics

1 answer:

lukranit [14]2 years ago

4 0

Answer:

Ann=£38

Bill=£171

Chloe=£95

Dave=£95

Step-by-step explanation:

By the information provided we can create an equation:

$2x+9x+2*5x=399\\11x+10x=399\\21x=399\\x=19$

Now that we know x, we multiply it correspondingly.

For Ann -> 2*19=38

Bill -> 9*19= 171

Chloe and Dave have the exact same amount each (because they both have 2.5 times the amount Ann's money)

Chloe/ Dave -> 5*19=95 each.

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The sum of an infinite geometric sequence is seven times the value of its first term.

Radda [10]

Answer:

a). r = $\frac{6}{7}$

b). At least 5 terms should be added.

Step-by-step explanation:

Formula representing sum of infinite geometric sequence is,

$S_{\inf}=\frac{a}{1-r}$

Where a = first term of the sequence

r = common ratio

a). If the sum is seven times the value of its first term.

$7a=\frac{a}{1-r}$

$7=\frac{1}{1-r}$

7(1 - r) = 1

7 - 7r = 1

7r = 7 - 1

7r = 6

r = $\frac{6}{7}$

b). Since sum of n terms of the geometric sequence is given by,

$S_{n}=\frac{a(1-r^{n})}{1-r}$

If the sum of n terms of this sequence is more than half the value of the infinite sum.

$\frac{a[1-(\frac{6}{7})^{n}]}{1-\frac{6}{7}}$ > $\frac{7a}{2}$

$\frac{1-(\frac{6}{7})^{n}}{1-\frac{6}{7}}> \frac{7}{2}$

$\frac{1-(\frac{6}{7})^{n}}{\frac{1}{7}}> \frac{7}{2}$

$1-(\frac{6}{7})^{n}> \frac{7}{2}\times \frac{1}{7}$

$1-(\frac{6}{7})^{n}> \frac{1}{2}$

$-(\frac{6}{7})^{n}> -\frac{1}{2}$

$(\frac{6}{7})^{n}< \frac{1}{2}$

$(0.85714)^{n}< (0.5)$

n[log(0.85714)] < log(0.5)

-n(0.06695) < -0.30102

n > $\frac{0.30102}{0.06695}$

n > 4.496

n > 4.5

Therefore, at least 5 terms of the sequence should be added.

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2 years ago

there is 4/5 lb of sand in the class science supplies. if one scoop of sand weighs 1/20 lb, how many scoops of sand can maria ge

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2 years ago

Arjay, Dorothy, Melissa, and Gray live in the same city. Arjay and Dorothy live 2 miles from each other. Dorothy and Melissa liv

GarryVolchara [31]

Dorothy and Gray live 8 miles from each other.

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2 years ago

) The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students

katen-ka-za [31]

Answer:

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

Step-by-step explanation:

We are in posessions of the sample's standard deviation. So we use the student's t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}$ ). So we have T = 2.0930