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3 years ago

Write 3 equivalent ratios for 3/5

Mathematics

2 answers:

Doss [256]3 years ago

8 0

6/10
9/15
12/20 it should be it :)

pashok25 [27]3 years ago

4 0

I believe these might answer your question
6/10
9/15
12/20

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What's is the slope of this line? Need this ASAP

Basile [38]

Answer:

$\displaystyle m=2$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Reading a Cartesian Plane

Coordinates (x, y)

Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Find points from graph.

Point (0, 3)

Point (1, 5)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [Slope Formula]: $\displaystyle m=\frac{5-3}{1-0}$
[Slope] Subtract: $\displaystyle m=\frac{2}{1}$
[Slope] Divide: $\displaystyle m=2$

5 0

3 years ago

Mark is 15 years younger than Ashley. The sum of their ages is 55. Write a system of 2 equations to determine Mark (M) age and A

skelet666 [1.2K]

55=2a-15 or 55=a+(a-15)

6 0

3 years ago

Cos15°°- Sin 15°= 1/√2

ololo11 [35]

<h2>Refer this attachment</h2>

Hope it's help you

4 0

3 years ago

A sample of blood pressure measurements is taken for a group of adults, and those values (mm Hg) are listed below. The values

Slav-nsk [51]

Answer:

Systolic on right

$\hat{CV} =\frac{18.68}{127.5}=0.147$

Systolic on left

$\hat{CV} =\frac{12.65}{74.2}=0.170$

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

Step-by-step explanation:

Assuming the following data:

Systolic (#'s on right) Diastolic (#'s on left)

117; 80

126; 77

158; 76

96; 51

157; 90

122; 89

116; 60

134; 64

127; 72

122; 83

The coefficient of variation is defined as " a statistical measure of the dispersion of data points in a data series around the mean" and is defined as:

$CV= \frac{\sigma}{\mu}$

And the best estimator is $\hat {CV} =\frac{s}{\bar x}$

Systolic on right

We can calculate the mean and deviation with the following formulas:

[te]\bar x = \frac{\sum_{i=1}^n X_i}{n}[/tex]

$s= \frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}$

For this case we have the following values:

$\bar x = 127.5, s= 18.68$

So then the coeffcient of variation is given by:

$\hat{CV} =\frac{18.68}{127.5}=0.147$

Systolic on left

For this case we have the following values:

$\bar x = 74.2 s= 12.65$

So then the coeffcient of variation is given by:

$\hat{CV} =\frac{12.65}{74.2}=0.170$

So for this case we have more variation for the data of systolic on left compared to the data systolic on right but the difference is not big since 0.170-0.147 = 0.023.

4 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago