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3 years ago

4. There are 10 pupils in a class. Half of them are

Mathematics

1 answer:

Vlad1618 [11]3 years ago

4 0

Answer:

hello :)

no quite sure but I think it is:

50/100

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in class you earned 10 participation points one day, -14 the next, and 6 the following day. how many points did you earn that we

Katena32 [7]

Answer:

2 points

Step-by-step explanation:

10-14=-4

-4+6=2 or

10+6=16

16-14=2

8 0

3 years ago

Describe the difference in how the pie chart and the Pareto chart show patterns in the data. Choose the correct answer below.

mars1129 [50]

Answer:

A. The pie chart shows the percentages as parts of the whole. The Pareto chart shows the rankings of the seasons.

Step-by-step explanation:

hope this helps

5 0

3 years ago

Help please thank you.

Shkiper50 [21]

$3 \sqrt{5} + 6 \sqrt{9} \sqrt{5}$
$3 \sqrt{5}+6 \sqrt{45}$
$3 \sqrt{5}+6*3 \sqrt{5}$
$3 \sqrt{5} + 18 \sqrt{5} = 21 \sqrt{5}$

5 0

3 years ago

Writing prompt about math

Nookie1986 [14]

Answer:

y=(3x+4)(x+2)

Step-by-step explanation:

Once you distibute everything you will get:

3x^2

which will give you your intitialequation: y=3x^2+10x+8.

The first equation you put can not be fcatored.

7 0

3 years ago

Suppose the national mean SAT score in mathematics was 510. In a random sample of 60 graduates from Stevens High, the mean SAT s

zalisa [80]

Answer:

We accept the alternate hypothesis and conclude that mean SAT score for Stevens High graduates is not the same as the national average.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 510

Sample mean, $\bar{x}$ = 502

Sample size, n = 60

Sample standard deviation, s = 30

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 510\\H_A: \mu \neq 510$

We use Two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }$ Putting all the values, we have

$t_{stat} = \displaystyle\frac{502- 510}{\frac{30}{\sqrt{60}} } = -2.0655$

Now,

$t_{critical} \text{ at 0.05 level of significance, 59 degree of freedom } = \pm 1.671$

Since,

$|t_{stat}| < |t_{critical}|$

We reject the null hypothesis and fail to accept it.

We accept the alternate hypothesis and conclude that mean SAT score for Stevens High graduates is not the same as the national average.

7 0

3 years ago