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Ymorist [56]
3 years ago
5

−12(14−23)= what is the answer to this

Mathematics
2 answers:
fredd [130]3 years ago
5 0
108 is the answer I’m pretty sure
LenaWriter [7]3 years ago
3 0

Answer:

108

Step-by-step explanation:

PEMDAS:

Parenthesis

Exponents

Multiplication and Divison

Addition and Subtraction

First Parenthesis first so 14-23 is -9

So the equation looks like this:

-12(-9)

-12 times -9 is 108

Hope this helps!

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vitfil [10]
2x-2y
I couldn’t tell if the first figure was a one or a L or i so I solved it as a 1
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2 years ago
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2.465/ 2.5 but I need to show my work.
Ivahew [28]

Answer:

0.986

Step-by-step explanation:

2.5 times 0.986=2.465

8 0
2 years ago
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
The price of an item yesterday was $160 . Today, the price fell to 104 . Find the percentage decrease.
babymother [125]
The percentage drop would be 35%
6 0
3 years ago
Write the equation of a linear function for which f=(-4)=0 and f=(0)=-2
densk [106]

Answer:

y = -0.5x - 2

Step-by-step explanation:

equation: f(x) = mx + b

f(-4) : -4m + b = 0  ...(1)

f(0):   0m + b = -2            b = -2   ... plug in (1)

-4m - 2 = 0

m = - 1/2 = - 0.5

linear equation: y = -0.5x - 2

6 0
3 years ago
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