An account earns simple interest. $600 at 5% for 2 years

I'm thinking of a fraction that is equivalent to . The numerator is less than the denominator." What fraction is thinking of?

Vesna [10]

Answer:

Proper fraction

Step-by-step explanation:

If the numerator is less than the denominator , then it is proper fraction.

Example : $\frac{5}{8}, \frac{124}{1000}$

8 0

3 years ago

How much of a spice that is 3% salt should be added to 175 ounces of a spice that is 6% salt in order to make a spice that is 5%

kozerog [31]

<h3>Answer:</h3>

87.5 ounces

<h3>Explanation:</h3>

There's a cute "X-diagram" tool that can be used for solving mixture problems of all kinds. (See the attachment.) The numbers on the left represent the percentages of the constituents of the mix. The number in the center is the desired percentage in the resultant mix. The numbers on the right in the X are the differences along the diagonals. (6-5=1, 5-3=2)

These are also the proportions of the constituents in the final mix. That is, there are 2 parts of spice that is 6% salt for each 1 part of spice that is 3% salt. So, we need half as much of the latter as the former.

Half of 175 ounces is 87.5 ounces, the amount of 3% spice that is needed.

_____

If you really want to write an equation, you can let x represent the amount of 3% spice needed. Then the amount of salt in the mix is ...

... 175×6% +x×3% = (175+x)×5%

Dropping the percent symbols and eliminating parentheses, we have

... 1050 +3x = 875 +5x

... (1050 -875) = (5x -3x) . . . . . add -875-2x

... 175/2 = x = 87.5 . . . . . . . . . . divide by the coefficient of x

5 0

3 years ago

Read 2 more answers

Y’all please help me everybody is finishing the test and I’m still here ☝️but don’t send me a link ✋

sukhopar [10]

Answer:

0

Negative

Un-defined

Things you need to know:

Any line parallel to the x axis is 0

any line parallel to the y axis is infinite/un-defined

Any line that is slanted in that way shown in the picture is negative vice versa the other way it is slanted will be positive.

8 0

3 years ago

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At a store 20 feet of fencing cost $36 at the rate how much will 15 feet of fencing cost

maria [59]

Answer:

15 feet of fencing costs $27

Step-by-step explanation:

We know 15/20 = 3/4 so...

36 x 3/4 = 27

Another way:

36 divided by 20 = 1.8

1.8 x 15 = 27

8 0

3 years ago

Read 2 more answers

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

C

b

C

c

C

d

A

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago