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3 years ago

My index finger is 3.2in long... so in cm is it 8.128cm?

Mathematics

2 answers:

kiruha [24]3 years ago

8 0

Yes that is correct..

BartSMP [9]3 years ago

4 0

There are 2.54 centimeters in 1 inch. But 8.128 centimeters in 3.2 inches. So, yes you are correct.

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Match each equation to its equivalent equation in slope-intercept form.

Bumek [7]

1.) y+6=3(x+2) C
y+6=3x+6
y=3x+6-6
y=3x+6
2.) y=1/2(x+8)-2 B
y=1/2x+4-2
y=1/2x+2
3.) y+1=1(x-3) E
y+1=x-3
y=x-3-1
y=x-4
4.) -4x+y=-2 A
y=-4x-2
5.) 2x-4y=-4 F
-4y=-2x-4
y=-2/-4x-4/-4
y=2/4x+4/4
y=1/2x+1
6.) 2x+4y=8 D
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6 0

3 years ago

The image shows the linear graph for a family that hired a babysitter for a day. The family gave the babysitter some money up fr

AleksandrR [38]

Answer:

I think it's A and C

7 0

3 years ago

Read 2 more answers

What is - 6 5p - p when you simplify it

Darina [25.2K]

5p-p=4p. The question doesn't make sense because ther is no sign after the minus 6. If it is an X it would be -24p. If it is an add it would be -6+4p. If it was a - it would be -6-4p.

6 0

3 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

86 teams enter the race .if the race begins with 16 dogs on each team how many dogs will enter the race

brilliants [131]

Answer:

1376 dogs

Step-by-step explanation:

This is just a multiplication problem if we convert the words to mathematical formulas.

There are 86 teams: $a_1,a_2,...a_8_5,a_8_6$ . Each team has 16 dogs: $a_1=16,a_2=16,...a_8_5=16,a_8_6=16$ .