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3 years ago

5

Meg has 24 feet of ribbon that she wants to divide into 3/5 of a foot pieces. How many 3/5 of a foot pieces are there in 24 feet

?

2 answers:

andre [41]3 years ago

8 0

14 2/5 segments

24 divided by 3/5 doesn’t work so you have to multiply them
24 x 3/5 = 14 2/5

g100num [7]3 years ago

5 0

Answer:

40

Step-by-step explanation:

24 feet ÷3/5 =

24/1 × 5/3 =

40

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Use the graph below to determine a1 and d for the sequence. graphed sequence showing point 1, negative 10, point 2, negative 7,

Sonja [21]

Answer:

$a_1=-10$

$d=3$

Step-by-step explanation:

we know that

In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant, and this constant is called the common difference (d)

In this problem we have the ordered pairs

$(1,-10),(2,-7),(3,-4),(4,-1),(5,2),(6,5)$

Let

$a_1=-10\\a_2=-7\\a_3=-4\\a_4=-1\\a_5=2\\a_6=5$

Find the difference between one term and the next

$a_2-a_1=-7-(-10)=3$

$a_3-a_2=-4-(-7)=3$

$a_4-a_3=-1-(-4)=3$

$a_5-a_4=2-(-1)=3$

$a_6-a_5=5-2=3$

The difference between one term and the next is a constant

This constant is the common difference

so

The sequence graphed is an Arithmetic Sequence

therefore

The first term is $a_1=-10$

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4 0

3 years ago

Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v

Veronika [31]

Answer:

$\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = \$71$

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = \$ 22.35$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 20 - 1 = 19

From the t-table at α = 0.025 and DoF = 19

t-score = 2.093

So, the margin of error is

$MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\$

So the required 95% confidence interval is

$\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

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3 years ago