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LiRa [457]
2 years ago
12

Plz help I need help plz

Mathematics
2 answers:
marysya [2.9K]2 years ago
8 0
The answer is 10/3. Hope this helps
faltersainse [42]2 years ago
5 0

➠ \:\frac{10}{3}\: will \:be\:the\: answer.\\

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Solve for t 6t=30<br><br><br><br> math
lesya [120]

⟹\sf \: 6t = 30

⟹\sf \: Substitute \: the \: value \: of \: t \: as \: 30/6 \: in \: the \: above \: equation.

\sf \: 6t = 30 \\ \sf \: t =  \frac{30}{6}  \\ \sf \: t = 5

Answer ⟶ \boxed{\bf{t = 5}}

4 0
2 years ago
Read 2 more answers
Write an equation in point-slope form of the line that passes through (-1, – 4) with slope -2.
Rom4ik [11]
Y+4=-2(x+1)

This is the point slope form (y-y1) =m(x-x1)

I hope it will helped!
4 0
3 years ago
103,727,495 in a word form
3241004551 [841]
103,727,495 in word form is: one hundred three million, seven hundred twenty-seven thousand, four hundred ninety-five.
3 0
3 years ago
Which combination of integers can be used to generate the Pythagorean triple (5,12,13)
Kitty [74]

Answer:

x=3 and y=2

Step-by-step explanation:

The pythagorean triples are generated by two integrers x and y that can be found by solving the following system of equations:

\left \{ {{x^{2}-y^{2}=5}\atop {2xy=12}} \atop {x^{2}+y^{2}=13}}\right.

Solve the system of equations, and we get that the solution is x=3 and y=2.

Therefore, the combination of integrers that ca be used to generate the pythagorea triple are: x=3 and y=2

3 0
3 years ago
Read 2 more answers
Suppose that a random sample of 10 adults has a mean score of 62 on a standardized personality test, with a standard deviation o
aniked [119]

Answer:

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645*\frac{4}{\sqrt{10}} = 2.08

The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92

The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

5 0
2 years ago
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