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dybincka [34]
3 years ago
5

Factor by first factoring out the greatest common 1actor 15x 35x-30 15x-35x-30​

Mathematics
2 answers:
Bingel [31]3 years ago
6 0

Answer:

5(3x²-7x-2)

Step-by-step explanation:

15x²-35x-30

5(3x²-7x-2)

Lostsunrise [7]3 years ago
5 0

Answer:

5(3x+2)(x−3)

Step-by-step explanation:

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3 years ago
Choose the correct set of y-values for the function y = -2x when x = -2, 0, 2.
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3 0
3 years ago
Which is a counterexample that disproves the conjecture? After completing several multiplication problems, a student concludes t
Dahasolnce [82]
When you make the product of a binomial of the kind x + a times other binomial that is of the kind x - a, you obtain another binomial (not a trinomial), so any example with that form will be a counterexample that disproves the conjecture:

(x + a) * (x - a) = x^2 - a^2

For example, (x +3) * (x - 3) = x^2 - 9. So, not always the product of two binomials is a trinomial.
5 0
3 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
I need help 5-7 please
OlgaM077 [116]
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6 0
3 years ago
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