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Mathematics

2 answers:

Rudiy273 years ago

7 0

Answer:

wouldn't it be 5m....??

Salsk061 [2.6K]3 years ago

6 0

Non factorable with rational numbers

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Find the zeros of the function f(x)=2x^2-7x-30

ivanzaharov [21]

$2x^2-7x-30=0 \\ \\
a=2 \\ b=-7 \\ c=-30 \\
b^2-4ac=(-7)^2-4 \times 2 \times (-30)=49+240=289 \\ \\
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-7) \pm \sqrt{289}}{2 \times 2}=\frac{7 \pm 17}{4} \\
x=\frac{7-17}{4} \ \lor \ x=\frac{7+17}{4} \\
x=\frac{-10}{4} \ \lor \ x=\frac{24}{4} \\
x=-\frac{5}{2} \ \lor \ x=6 \\
\boxed{x=-2.5 \hbox{ or } x=6}$

6 0

3 years ago

Solve for x: 5|2x − 2| + 8 = 18.

Nitella [24]

5|2x-2|+8=18
5|2x+2|=10
|2x+2|=2
by removing the absolute and but + or -
(2x+2) = 2 & (2x+2)=-2
so , x = 0 & x=-2
so it is the second answer

8 0

3 years ago

An article reported on the results of an experiment in which half of the individuals in a group of 60 postmenopausal overweight

slavikrds [6]

Answer:

Step-by-step explanation:

From the given information:

Let's first compute the null and alternative hypothesis

Null hypothesis:

$\mathbf{H_o: \mu_1-\mu_2=1}$

Alternative hypothesis:

$H_a :\mu_1 -\mu_2 > 1$

The number of samples is half in a group of 60

i.e

$n_1=n_2 = 30$

the sample mean for sample 1 $\overline x_1$ = 5.7

the standard deviation for sample 1 $s_1$ = 3.1

the sample mean for sample 2 $\overline x_2$ = 3.9

the standard deviation for sample 2 $s_2$ = 2.7

degree of freedom for this test can be computed by using the formula:

$df = \dfrac{\begin {pmatrix} \dfrac{s_1^2}{n_1} + \dfrac{s^2_2}{n_2} \end {pmatrix}^2 } {\dfrac{ (\dfrac{s_1^2}{n_1}^2)}{ n_1-1} + \dfrac{ (\dfrac{s_2^2}{n_2}^2)}{ n_2-1} }$

$df = \dfrac{\begin {pmatrix} \dfrac{3.1^2}{30} + \dfrac{2.7^2}{30} \end {pmatrix}^2 } {\dfrac{ \begin {pmatrix} \dfrac{3.1^2}{30} \end {pmatrix}^2}{ 30-1} + \dfrac{ \begin {pmatrix} \dfrac{2.7^2}{30} \end {pmatrix}^2}{ 30-1} }}$