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3 years ago

Find the samllest value of K when : 280k is a perfect square 882 is a cube

1 answer:

5 0

280 = 7 x 2^3 x 5
882 = 7^2 x 2 x 3^2

For 280k to be perfect square , their powers need to be even
For 882 to be a cube, their powers need to be in multiple of 3,

To satisfy both parties,
K = 7 x 2^5 x 3^4 x 5^3 = 2268000

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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

The number of vibrations n n per second of a nylon guitar string varies directly with the square root of the tension T T and inv

emmasim [6.3K]

Answer: $T=40.96\ KgF$

Step-by-step explanation:

We know that:

$n$ : The number of vibrations per second of the nylon guitar string.

$T$ : Tension.

$L$ : The length of the string.

Since $n$ varies directly with the square root of the $T$ and inversely with $L$ , the equation has the following form:

$n=k*\frac{\sqrt{T} }{L}$

Where "k" is the constant of variation.

Knowing that when $n=15$ and $L=0.6$ , $T=256$ , we can find the value of "k":

$n=k*\frac{\sqrt{T} }{L}\\\\L*n=k\sqrt{T}\\\\\frac{L*n}{\sqrt{T}}=k\\\\k=\frac{(0.6)(15)}{\sqrt{256}}\\\\k=0.5625$

Finally, in order to find the tension when the length is 0.3 meters and the number of vibrations is 12, you need to substitute these values and the value of "k" into $n=k*\frac{\sqrt{T} }{L}$ and solve for $T$ :

$12=(0.5625)\frac{\sqrt{T} }{0.3}\\\\\frac{12(0.3)}{(0.5625)}=\sqrt{T}\\\\(6.4)^2=T\\\\T=40.96\ KgF$

8 0

3 years ago

What is the lcm of 7,6,8

saw5 [17]

7 = 1 x 7
6=2x3
8=2x2x2
LCM= 7x2x2x2x3= 8 x 21 = 168

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3 years ago