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lys-0071 [83]
3 years ago
14

I need help bad! I need help with the last question!

Mathematics
2 answers:
Gala2k [10]3 years ago
6 0
6 is the answer to the last question
MAVERICK [17]3 years ago
5 0
Use the fact that 1 axe head = 3.20 trade value.
Let N denote number of axe heads needed to trade for a trade gun.
3.20×N=120.00
》 N=120/3.2 which will give number of axe heads needed. Round to the nearest ones place if you must.
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Find the square root of 3364 by prime factorization method
olganol [36]

3364

= 2^2 * 29^2

So

√3364 = √(2^2 * 29^2) = 2 * 29 = 58

3 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!!<br><br> Simplify.
NNADVOKAT [17]

Answer:  \frac{2x(5x + 1)}{4x - 1}

<u>Step-by-step explanation:</u>

\frac{4x^{3}-12x^{2}}{4x^{2}+7x - 2} ÷ \frac{2x^{2}-6x}{5x^{2}+11x + 2}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} ÷ \frac{2x(x - 3)}{(5x + 1)(x + 2)}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} x \frac{(5x + 1)(x + 2)}{2x(x - 3)}

= \frac{2x(5x + 1)}{4x - 1}

8 0
3 years ago
Has any see alexalunathis moring
Irina18 [472]

Answer:

no ill tell u if i see her homedawgidawg

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

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

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
A veterinarian is going to administer a medication which has a 3.2 liquid to drug
Margaret [11]
The he gives some drugs
4 0
3 years ago
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