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4 years ago

How to factor 9x^2-25

Mathematics

2 answers:

noname [10]4 years ago

7 0

9x^2 - 25 = (3x-5)(3x+5)

sveticcg [70]4 years ago

4 0

Since the 9 and 25 are both perfect squares and it is a difference (subtraction), you have a difference of perfect squares. This always factors to the form
(a+b)(a-b).
(3x-5)(3x+5) is the answer

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Semenov [28]

Bro what are you saying my guy

7 0

3 years ago

Please help! Btw the pencil is wrong answers. Thank you so much!!!

VLD [36.1K]

Answer:

Below are explanations!

Step-by-step explanation:

6) 5^4/5^2 means the ^4 and ^2 needs to subtract from the 5s. This results in 5^2, which equals 25.
7) The 9 as the denominator will cross with the first nine above. This leaves as follows: 9^3 x 9^2 x 9^6 (I don't know if it is a 6 or 0). You may solve from there. Just add all the exponents up.
8)^-6 will be multiplied by ^6, which is ^-36. Therefore, the answer is 5^-36.
The other questions, please, remember to read carefully.

6 0

3 years ago

Given two dependent random samples with the following results:

skad [1K]

Answer:

We Fail to reject Null Hypothesis.

Step-by-step explanation:

We are given two dependent random samples;

Population 1 Population 2 d = Population 1 entry-Population 2 entry $d^{2}$

52 61 52 - 61 = -9 81

81 71 81 - 71 = 10 100

72 82 72 - 82 = -10 100

67 62 67 - 62 = 5 25

73 79 73 - 79 = -6 36

73 77 73 - 77 = -4 16

76 71 76 - 71 = 5 25

83 85 83 - 85 = -2 4

∑ $d^{2}$ = 387

Now, Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ or $\mu_d = 0$

Alternate Hypothesis, $H_1$ : $\mu_1 \neq \mu_2$ or $\mu_d \neq 0$

The test statistics for paired data is given by;

$\frac{dbar - \mu_d}{\frac{s_d}{\sqrt{n} } }$ follows $t_n_-_1$

where, $dbar$ = Mean of Population 2 data - Mean of Population 1 data

Mean of Population 1 data = $\frac{52+81+72+67+73+73+76+83}{8}$ = 72.125

Mean of Population 2 data = $\frac{61+71+82+62+79+77+71+85}{8}$ = 73.5

Hence, $dbar$ = 73.5 - 72.125 = 1.375

$s_d$ = Standard Deviation of paired data = $\sqrt{\frac{\sum d^{2} - n*(dbar)^{2} }{n-1} }$ = $\sqrt{\frac{387 - 8*(1.375)^{2} }{8-1} }$

= 7.3

n = no. of observations = 8

So, Test Statistics = $\frac{1.375 - 0}{\frac{7.3}{\sqrt{8} } }$ follows $t_7$

= 0.533

Now, if the critical value of t at 7 degree of freedom is less than the test statistics then we will reject null hypothesis.

At 0.01 level of significance, the t table gives a critical value of 3.499. Since our test statistics is less than the critical value we will not reject null hypothesis and conclude that there is no significant difference between the two population means.