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dem82 [27]
3 years ago
11

What is 60 as a ratio

Mathematics
1 answer:
Pani-rosa [81]3 years ago
6 0

Answer:

the answer is 60/100 60%

Step-by-step explanation:

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JIH= QRP
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MAXImum [283]

Answer: \frac{1}{b^6}   or b^{-6}  

Step-by-step explanation:

\frac{b^-2+2}{b^4+2}   Add the exponent two to both sides.  After adding -2 and     2 you get 0 and any number raised to the zero power is one. And 4 plus two is 6

\frac{b^0}{b^6}   = \frac{1}{b^6}

3 0
3 years ago
Someone please help me out i don´t understand this question <br> ill give you 21 points
suter [353]

Answer:

0.67

Step-by-step explanation:

Formula: k = y/x

k = 2/3

k = 0.66666666666

k = 4/6

k= 0.66666666666

k = 6/9

k = 0.66666666666

k = 8/12

Round 0.67

Hence, answer = 0.67

[RevyBreeze]

8 0
2 years ago
Read 2 more answers
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
3 years ago
Where do you place the parentheses in the expression 4+3^2•5-2 to equal 31
Pani-rosa [81]

4+3^2•5-2

4+3^2\cdot(5-2)=4+9\cdot3=4+27=31

Used PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

8 0
3 years ago
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