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Svet_ta [14]
3 years ago
11

Q1 Express in the form 1.0 4:12

Mathematics
1 answer:
kherson [118]3 years ago
3 0

Answer:

Step-by-step explanation:

4:12= 1:3

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On a fruit farm 1/3 of the trees are apple trees and 1/5 are lemon trees. The rest are fig trees. There are 90 trees in all.
Vaselesa [24]

Answer:

I'm not sure what the question was but I've worked out all the three types of trees

Step-by-step explanation:

⅓ + ⅕ = 8/15 because:

×5 ×3

5/15 + 3/15 = 8/15

therefore 7/15 is fig trees

90÷15 =6

6×5 = 30 apple trees

6×3 = 18 lemon trees

6×7 = 42 fig trees

7 0
2 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
I’m having a hard time finding the answer, any help is appreciated (:
kogti [31]

Answer:

True.

Step-by-step explanation:

It fails the vertical line test. The vertical line test checks to see if any inputs (x values) have more than one output. In this case, they do.

3 0
3 years ago
How do we solve this one​
Whitepunk [10]

Answer:

how should I answer it?

Step-by-step explanation:

elimination,substitution, etc

answer will be in comments

7 0
2 years ago
Substract the polynomials and simplify (-3r2-2r-4)-(9r2+7r+2)
mina [271]

Answer:

-12r2-9r-6

Step-by-step explanation:

-distribute the subtraction symbol  to the second group of numbers

-(9r2+7r+2) = (-9r2-7r-2)

-then combine each term together (-3r2-2r-4)+(-9r2-7r-2)

= -12r2-9r-6

HOPE THIS HELPS!!

6 0
3 years ago
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