Simplify 10^-8 .

Please help me solve an angle in right triangle

densk [106]

Answer:

8

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A triangle has ∠1 and ∠2 as remote interior angles with respect to exterior angle ∠3. Given that m∠1 = 50 and m∠2 = 70, Alicia r

11111nata11111 [884]

Answer:

No, 60 is the third interior angle's mesure, not the exterior ∠3.

Step-by-step explanation:

Say that ∠1=∠A=50 and ∠2=∠B=70. The remote interior angles are the angles that are not touvhing the exterior angle. So ∠3 would be the the exterior angle to ∠C. ∠C=60 as the sum of the interior angles is 180.

The exterior angle is the angle formed by extending one of the triangle's sides. A straight line has the angle 180, so∠3=180-∠C=180-60=120.

4 0

3 years ago

How does multiplying a vector by a scalar value of -pi change the vector?

Dmitriy789 [7]

I believe the answer to your question is A

6 0

4 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

Prove that the number A= [tex]20^{8^{2014} } }+113 is composite

Drupady [299]

Cheese proof:

We can just prove that it is divisible by 3, which means that it is composite. We can use modular exponentiation, where if $a \equiv b \pmod{n}$ then $a^x \equiv b^x \pmod{n}$ . In this case, ${-1}^{8^{2014}} \equiv 20^{8^{2014}} \pmod{3}$ . This is much easier to calculate! Since $8^{2014}$ is even, $-1^{8^{2014}}=1$ , meaning that now we only need to prove that $0\equiv(1+113) \pmod{3}$ , which is obviously true.

8 0

2 years ago