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exis [7]
3 years ago
5

Find the area please and will make brainliest

Mathematics
1 answer:
puteri [66]3 years ago
4 0
To find the area of the trapezoid, we only need the quantities a1 (long base), a2(short base) and h(height).

Area=(1/2)(a1+a2)*h
=(1/2)(9.9+4.7)(5.6)
=40.88 mm ²

Not sure what's expected to give for "type".  Perhaps trapezoid or trapezium.

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Substitution method<br> x=2y<br> 2x+5y=4
Dvinal [7]

Answer:

y=4/9

Step-by-step explanation:

Substitute 2y into the x

2(2y)+5y=4

4y+5y=4

9y=4

y=4/9

6 0
2 years ago
Any gírl here<br>.<br>.<br>.<br>.<br>let's chàt​
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uhm no stop this is an app for people and work so please go

3 0
2 years ago
An economy package of cups has 460 green cups. If the green cups are 20% of the total package, how many cups are in the package?
Oksi-84 [34.3K]
2300 cups because if 460 is 20%, multiply the 460 x 5 to get 100%
3 0
2 years ago
John runs 4 miles in 30 minutes. at the same rate, how many miles would he run in 48 minutes?
dimulka [17.4K]
Make a ratio, then cross multiply.

4/30=x/48

30x=4x48
30x=192
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3 0
3 years ago
A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec
Ket [755]

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

(a)

The null hypothesis is:

<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

For 5% level of significance the two tailed critical value of <em>z</em> is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct the 95% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

3 0
2 years ago
Read 2 more answers
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