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2 years ago

What would happen if you subtracted the equation for x from the equation for 10x?

Mathematics

1 answer:

vladimir1956 [14]2 years ago

6 0

Answer:

9 x

Step-by-step explanation:

If you have 10 x and you subtract one, you get left with 9 x.

Think of it as objects. You have 10 apples and someone takes an apple, what do you have left? 9 x = 3

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3 years ago

A stack of playing cards contains 4 jacks, 5 queens, 3 kings, and 3 aces. two cards will be randomly selected from the stack. wh

Kipish [7]

Answer:

1/9.

Step-by-step explanation:

There is a total of 15 cards in the stack.

Prob( Queen is chosen) = 5/15 = 1/3.

The probability of a second queen being chosen is also 1/3

Required probability = 1/3 * 1/3 = 1/9.

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3 years ago

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3 years ago

Georgia has a budget of $8 for a new notebook. She wants to spend within $5 of her budget, so she uses the equation 5 = |8 – x|

siniylev [52]

Answer:

Minimum: 3

Maximum: 13

Step-by-step explanation:

Georgia uses the equation

5 = |8 – x| to find the maximum and minimum values.

We solve the equation for x.

$5 = |8 - x|$

By the definition of the absolute value function, we must have:

$- 5 = 8 - x \: or \: 5 = 8 - x$

We subtract 8 from both sides to get:

$- 5 - 8 = - x \: or \: 5 - 8 = - x$

$- 13 = - x \: or \: - 3 = - x$

This simplifies to:

$13 = x \: or \: 3 = x$

Therefore the minimum value is 3 and maximum is 13

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].

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3 years ago