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3 years ago

PLEASEE HELP!! Find the area of the kite.

Mathematics

1 answer:

aev [14]3 years ago

6 0

Answer:

Step-by-step explanation:

Area of kite is Area of triangles. Area is 6+3=9

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Hi this is urgent. i have a few algebra models that need to be done in 10 days. If anyone one can do them corretly with an 80 of

GREYUIT [131]

Answer:

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3 years ago

A.) 2 logbq + 8 logbt B.) 4 log x – 6 log (x + 2)

jasenka [17]

A) $2log_{b}q + 8 log_{b} t= \\ log_{b} q^{2} + log_{b} t^{8} = \\ log_{b} q^{2} t^{8}$
Answer: B )
b ) $4 log x - 6 log( x + 2 ) = log x^{4} - log ( x+2 )^{6}= \\ log \frac{ x^{4} }{(x+2) ^{6} }$
Answer: D ) None of those

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3 years ago

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Nimfa-mama [501]

$68.82
$68.82 - $5.32
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$60.50 / 2
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4 years ago

Read 2 more answers

I need help with math!!! time is running out!!! will mark brainliest!!!

sashaice [31]

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Answer B

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Answer D.

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3 years ago

The 2000 census "long form" asked the total 1999 income of the householder, the person in whose name the dwelling unit was owned

shutvik [7]

Answer:

The 99% confidence interval for the 1999 mean incomes of Middletown householders is ($33,013.5, $33,892.5).

Step-by-step explanation:

Of the 40,000 persons, 2621 householders reported their income.

The sample size of the study is, n = 2621.

As the sample size is large, i.e. n > 30 use z-distribution to compute the confidence interval.

The confidence interval for mean is:

$CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$

The mean of this sample is, $\bar x=\$33,453$

The standard deviation is, $s\approx\sigma=\$8,721$

The critical value of z for 99% confidence level is:

$z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58$

The 99% confidence interval for the 1999 mean incomes of Middletown householders is:

$CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\\=33453\pm2.58\times\frac{8721}{\sqrt{2621}}\\=33453\pm439.50\\=(33013.5, 33892.5)$

Thus, the 99% confidence interval for the 1999 mean incomes of Middletown householders is ($33,013.5, $33,892.5).

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3 years ago