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Nonamiya [84]
3 years ago
7

I need the answer ASAP! Thanks! ❤️

Mathematics
2 answers:
zubka84 [21]3 years ago
6 0
Your answer will be sting and positive
Yuliya22 [10]3 years ago
5 0
Strong and positive
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What is the answer to a,b, and c? Thanks!
Katarina [22]

The area of a triangle can be calculated as half the product of base length and height. We want the area to be no greater than 10 in².

a) (1/2)(4)(2x-3) ≤ 10

b) 4x -6 ≤ 10 . . . . . simplify

... 4x ≤ 16 . . . . . . . .add 6

... x ≤ 4 . . . . . . . . . divide by the coefficient of x

c) The maximum value of (2x -3) in is (2·4 -3) in = 5 in.

The triangle should be no more than 5 in high to have an area less than 10 in².

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3 years ago
Find the product of 2x^(2x2 + 3x + 4)
madam [21]
Answer- 4x⁶ + 6x⁵ + 8x⁴
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3 years ago
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70 POINTS!!!! PLEASE HELP I REALLY NEED HELP ESPECIALLY PART D AND F!!!!
sashaice [31]
Bear in mind that for y intercept x=0 and for x intercept y=0
5 0
3 years ago
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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
Solve y = x + 6 for x.
elena-s [515]

Answer:

C

Step-by-step explanation:

move the x over

-x+y=6

move the y over

-x = -y+6

times each number by a -

x=y-6

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2 years ago
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