can someone help me with this.( Rational and Irrational numbers. Convert the repeating decimal into fraction)

(3b-5)^2-c^2 how do I factor this using grouping?

klemol [59]

So you see it is the difference of 2 perfect squares
a^2-b^2=(a+b)(a-b)
so
(3b-5)^2-c^2=(3b-5+c)(3b-5-c)

3 0

3 years ago

An unknown number is half the product of 4 and 12 the number is ?

pochemuha

Let's use n to represent the unknown number:
n = 4*12/2
=48/2
= 24
The unknown number is 24

3 0

3 years ago

Do positive or negative messages have a greater effect on behavior? Forty-two subjects were randomly assigned to one of two trea

Alex_Xolod [135]

Answer:

We conclude that a negative message results in a lower mean score than positive message.

Step-by-step explanation:

We are given that Forty-two subjects were randomly assigned to one of two treatment groups, 21 per group.

The 21 subjects receiving the negative message had a mean score of 9.64 with standard deviation 3.43; the 21 subjects receiving the positive message had a mean score of 15.84 with standard deviation 8.65.

Let $\mu_1$ = population mean score for negative message

$\mu_2$ = population mean score for positive message

SO, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1\geq \mu_2$ {means that a negative message results in a higher or equal mean score than positive message}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1$ {means that a negative message results in a lower mean score than positive message}

The test statistics that will be used here is Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1+_n__2-2$

where, $\bar X_1$ = sample mean score for negative message = 9.64

$\bar X_2$ = sample mean score for positive message = 15.84

$s_1$ = sample standard deviation for negative message = 3.43

$s_2$ = sample standard deviation for positive message = 8.65

$n_1$ = sample of subjects receiving the negative message = 21

$n_2$ = sample of subjects receiving the positive message = 21

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(21-1)\times 3.43^{2}+(21-1)\times 8.65^{2} }{21+21-2} }$ = 6.58

So, the test statistics = $\frac{(9.64-15.84)-(0)}{6.58 \times \sqrt{\frac{1}{21}+\frac{1}{21} } }$ ~ $t_4_0$

= -3.053

Now at 0.05 significance level, the t table gives critical value of -1.684 at 40 degree of freedom for left-tailed test. Since our test statistics is less than the critical value of t as -3.053 < -1.684, 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 a negative message results in a lower mean score than positive message.

7 0

3 years ago

I need the answer for these. 1a: 1b: 1c:

Greeley [361]

$a.\\8:\dfrac{1}{2}\cdot3=8\cdot\dfrac{2}{1}\cdot3=8\cdot2\cdot3=16\cdot3=48\\\\b.\\3\sqrt{50-1}=3\sqrt{49}=3\cdot7=21\\\\c.\\\dfrac{(5+2)(-8)}{(-2)^3-3}=\dfrac{7(-8)}{-8-3}=\dfrac{-56}{-11}=\dfrac{56}{11}=5\dfrac{1}{11}$

7 0

3 years ago

Module 4 End Test Study

Snowcat [4.5K]

You'd do $500 x 0.15 to which gives you 75$ in tax. $500 + $75 = $575 selling price

5 0

3 years ago

can someone help me with this.( Rational and Irrational numbers. Convert the repeating decimal into fraction​)

can someone help me with this.( Rational and Irrational numbers. Convert the repeating decimal into fraction)