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3 years ago

Pls help me meememememmmememe

Mathematics

1 answer:

stellarik [79]3 years ago

3 0

Step-by-step explanation:

A= 4(pi)r^2

A=4(3.14159265359)13^2

A=2123.72

**I dont know if it's right or wrong.**

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David buys a new racing bike for £x he then sells it for £1584 making a loss of 12% find x

Sholpan [36]

Answer:

x=1800

Step-by-step explanation:

x-1584/x=0.12

x-1584=0.12x

x-0.12x=1584

0.88x=1584

x=1800

Hope this helps plz hit the crown :D

3 0

2 years ago

PLZ HELP ME ASAP!!!!!!!! THX!!!!!!!!!!!!!

topjm [15]

64/49

thats what I got but im not sure

5 0

3 years ago

Please help, will mark brainiest. 20+ POINTS

professor190 [17]

Answer:

2.5•2 = 5

Step-by-step explanation:

The resulting two-dimensional cross-section is the area of a square.

Area of a square is equal to:

$A=bh$

Having

$b = 2.5\ in\\h = 2\ in$

substitute

$A=2.5*2=5\ in^{2}$

4 0

2 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

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

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago

A doctor orders 20 grams of a 52% solution of a certain medicine. The pharmacist has only bottles of 40% solution and bottles of

Anvisha [2.4K]

Try, please, this:
1) Imagine, that pharmacist used 'x' gr. of 40% solution and 'y' of 70% solution.
Together (20 gr. of 52% sol.) it will be x+y=20.
2) From other side, the pharmacist combined 40x and 70y gramms, after obtaining together it is 52(x+y) gramms. In other words 40x+70y=52(x+y)
3)
$\left \{ {{x+y=20} \atop {40x+70y=52(x+y)}} \right. \ =\ \textgreater \ \ \left \{ {{y=8} \atop {x=12}} \right.$
Answer: 8 gr. of 70% solution and 12 gr. of 40%.

4 0

3 years ago

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