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3 years ago

Please help me whats 2+2-4/2?

Mathematics

2 answers:

marissa [1.9K]3 years ago

8 0

Answer:

Step-by-step explanation:

4/2=2

2+2=4

4-2=2

marta [7]3 years ago

5 0

Answer:

Step-by-step explanation:

(No step-by-step explanation)

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marusya05 [52]

By multiplying it to make it the same but different like 4/5×5/5 is 20/25 which is the same as 4/5

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4 years ago

I need helpful answers pls

Ivenika [448]

Answer:

X=5 the angle is 76 degrees

Step-by-step explanation:

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7 0

3 years ago

Can anyone help with this?

Sladkaya [172]

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3 years ago

100 points PLEASE ANSWER ASAP thank you so much Please show work

vivado [14]

Answer:

The mistake is that the equation on both sides is multiplied by 8. Both sides should be divided by 8.

Step-by-step explanation:

$8x^2-x-1=0$
$8x^2-x=1$ Divide both sides of the equation by 8
$x^2-\frac{1}{8} x=\frac{1}{8}$
$x^2-\frac{1}{8} x+?=\frac{1}{8} +?$ Add the same value to both side
$x^2-\frac{1}{8} x+\frac{1}{256} =\frac{1}{8} +\frac{1}{256}$
Use $a^2-2ab+b^2=(a-b)^2$
$(x-\frac{1}{16} )^2 = \frac{33}{256}$
$x-\frac{1}{16}=+-\frac{\sqrt{33} }{16}$
$x_{1} =\frac{-\sqrt{33}+1 }{16}$
$x_{2} =\frac{\sqrt{33}+1 }{16}$

5 0

2 years ago

Read 2 more answers

Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu

Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

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

where, $\bar X$ = sample mean procrastination score = 41

s = sample standard deviation = 6.89

n = sample of students = 69

$\mu$ = population mean estimate

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.9973 < $t_6_8$ < 1.9973) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.9973 & 1.9973 with P = 2.5%}

P(-1.9973 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.9973) = 0.95

P( $-1.9973 \times{\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ =[ $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ , $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ]

= [ $41-1.9973 \times{\frac{6.89}{\sqrt{69} } }$ , $41+1.9973 \times{\frac{6.89}{\sqrt{69} } }$ ]

= [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

5 0

3 years ago