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3 years ago

Change 45% to a decimal, then a fraction.

Mathematics

2 answers:

ratelena [41]3 years ago

5 0

Decimal Form: 0.45

Fraction Form: 45/100 or 9/20

Feliz [49]3 years ago

5 0

Answer:

In decimal = 0.45 and in fraction is, $\frac{45}{100}$ or $\frac{9}{20}$

Step-by-step explanation:

Convert 45% to a decimal.

We can write this 45% as:

$\frac{45}{100}$

=0.45

Now, convert it in a fraction.

45% in fraction.

$\frac{45}{100}=\frac{9}{20}$

In fraction: $\frac{45}{100}$ or $\frac{9}{20}$

Therefore,change 45% to a decimal, then a fraction are,

in decimal = 0.45 and in fraction is, $\frac{45}{100}$ or $\frac{9}{20}$

You might be interested in

The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 ent

SOVA2 [1]

Answer:

$t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744$

$p_v =P(t_{(47)}>2.744)=0.0043$

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=2.1$ represent the mean

$s=1.01$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =1.7$ represent the value that we want to test

$\alpha=0.02$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 1.7, the system of hypothesis would be:

Null hypothesis: $\mu \leq 1.7$

Alternative hypothesis: $\mu > 1.7$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=48-1=47$