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3 years ago

Write 16 3/5 as a decimal

Mathematics

2 answers:

zhuklara [117]3 years ago

3 0

16.6 is the answer. 16 is the whole number so u just divide 3 and 5 and get .6

gtnhenbr [62]3 years ago

3 0

16.6 hopefully this is right if not blame me

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WILL GIVE BRAINLY IF QUICK. 40 POINTS!!!!

sdas [7]

Answer:

x = $\sqrt{1}$

Step-by-step explanation:

(x + 6 )^2 = 49

get rid of the square , by square rooting the 49

x + 6 = $\sqrt{49}$

x = $\sqrt{49}$ -6

x = $\sqrt{1}$

$\sqrt{49}$ means + or - hence it could be -7 -6 = -1 ... or 7-6 = 1

therefore the answer is = or - 1

hence in terms of roots .. answer is :

$\sqrt{1}$

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3 years ago

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Which question can be answered by finding 212÷23? A. How many 23-meter long pieces of ribbon are needed to have a total of 212 m

SOVA2 [1]

The correct answer is the letter A because a division problem shows a total amount being broken into equal parts.

Choice a talks about how many parts it would take to get the total of 212 m of ribbon.

This choice includes the parts and total information.

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3 years ago

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In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

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3 years ago

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ankoles [38]

Answer:

Try this

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One bag of fertilizer = 4000 ft²

Number of bags of fertilizer needed = 7000 ÷ 4000 = 1.75 bags.

Since only full bags of fertilizer can be bought, 2 bags of fertilizer are needed.

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Answer: 2 bags of fertilizer
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3 years ago