Plz help me out ASAP

(2a^(3))^(-3) -------- (3b(-2))

Andrei [34K]

Hello,

$\dfrac{(2a^{3})^{-3}}{3b*(-2)}=-\dfrac{1}{(2a^3)^3*6b}=-\dfrac{1}{48a^9b}$

7 0

2 years ago

C+12=15 idk this pls help

Vedmedyk [2.9K]

Answer:

C = 3

Step-by-step explanation:

................

4 0

2 years ago

Read 2 more answers

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago

two telephone calls come into a switchboard at times that are uniformly distributed in a fixed one-hour period. assume that the

AleksandrR [38]

We wil assume a variable x to be the total number of calls received by the switchboard.

The question also says to assume that the calls were made independently.

Given:

Calls are independent.

Calls are uniformly distributed over a 1 hour period.

Ans (a). The calls are distributed uniformly over 1 hour, hence: (0, 1).

So we have,

f1(x1) = 1

f2(x2) = 1

X1 and X2 are considered to be independent of each other. Hence,

f(x1,x2) = f1 (x1) f2 (x2)

f(x1,x2) = 1 (1)

f(x1,x2) = 1

Thus,

P(X1 <= 0.5; X2 <= 0.5) = ∫0.50 ∫0.50 f(x1,x2) dx2 dx1

= ∫^0.5 0 ∫^0.5 0 (1) dx2 dx1

= ∫^0.5 0 (x2^0.5 0) dx1

= ∫^0.5 0 (0.5 - 0) dx1

= 0.5 ∫^0.5 0 dx1

= 0.5 (x1^0.5 0)

= 0.5 (0.5 - 0)

= 0.25

Therefore, the probability that the calls were received within the first 30 minutes or first half hour is 0.25.

Ans (b). Steps 1 and 2 are the same as the above answer.

Probability = [∫^11/12 0 ∫^x1 + 1/12 x1 1dx2 dx1] + [∫^1 1/12 ∫^x1 x1-1/12 1dx2 dx1

= [∫^11/12 0 (x2 ^x1+1/12 x1 dx1] + [∫^1 1/12 (x2 ^x1 x1-1/12 dx1)]

= [∫^11/12 0 (x1 + 1/12 -x1) dx1] + [∫^1 1/12 (x1 - x1 + 1/12) dx1]

= [(1 + x1/12 - 1) ^11/12 0] + [( 1 - 1 + x1/12) ^1 1/12]

= 11/144 + 11/144

= 0.1528

Therefore, the probability that the calls were received within five minutes of each other is 0.15.

#SPJ4

7 0

6 months ago

I dont understand how to do the 2x part.

kodGreya [7K]

Answer:

x=6 2/3

Step-by-step explanation:

Ratio of DEF:ABC=5/3

8*5/3=13 1/3

x=6 2/3

4 0

2 years ago