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4 years ago

Write thirteen and nine tenths in standard form

Mathematics

2 answers:

masha68 [24]4 years ago

5 0

Your aanswer is 13.9 because the 13 stays the same and in decimals .1 is a tenth so there for .9 is your tenth 13.9

creativ13 [48]4 years ago

5 0

Answer:

Therefore, 13.9

Step-by-step explanation:

Given : Thirteen and nine tenths.

To find : Write in standard form .

Solution : We have given thirteen and nine tenths.

Thirteen = 13

Nine tenths = $\frac{9}{10}$ .

13 + $\frac{9}{10}$ .

13 + 0.9

13.9

Therefore, 13.9

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The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systo

PIT_PIT [208]

Answer:

The estimate is

$52.02 < \mu < 55.78$

Step-by-step explanation:

From the question we are told that

The sample mean is $\ = x = 53.9$

The sample size is n = 24

The standard deviation is $\sigma = 5.6$

Given that the confidence level is 90% the level of significance is mathematically represented as

$\alpha = 100 - 90$

$\alpha = 10 \%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table.The value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} } = 1.645$

The reason we are obtaining critical value of $\frac{\alpha }{2}$ instead of $\alpha$ is because $\alpha$

represents the area under the normal curve where the confidence level interval ( $1 - \alpha$ ) did not cover which include both the left and right tail while

$\frac{\alpha }{2}$ is just the area of one tail which what we required to calculate the margin of error

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{ \sqrt{n} }$

substituting values

$E = 1.645 * \frac{5.6 }{ \sqrt{24} }$

$E = 1.880$

The estimate of how much the drug will lower a typical patient's systolic blood pressure(using a 90% confidence level) is mathematically represented as

$\= x - E < \mu < \= x + E$