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3 years ago

Why are halves a good choice for benchmark fractions for 1 1/3

Mathematics

1 answer:

Natalka [10]3 years ago

3 0

Answer and explanation:

Benchmark fractions are fractions that are used as references in measuring other fractions. They are easily estimated and so can be used in measuring more "specific" fractions such as 1/5, 7/9, 3/7, 1/3 etc. If I wanted to measure 1 1/3cm for instance using a calibrated ruler, having centimeter measurements, I would first find 1cm on the ruler and then find half of one centimeter. Seeing that half is bigger than 1/3 but close, I could then estimate 1/3 to be somewhere less than 1/2 but a bit close to it

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Solve the following expression when<br> m = 5 and d = 4<br> m2 + 2d + 8 + d

Tatiana [17]

If you use substitution the should be answer 30

3 0

3 years ago

By the Interior Angles Theorem, if Angle A is 25° and Angle B is greater than 51° but less than 57°, what are the possible measu

BARSIC [14]

98 < C < 104

180 - 25 - < C < 180 - 25 - 51
98 < C < 104

Measures the interior angles of a triangle sum to 180

3 0

3 years ago

Read 2 more answers

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

Natali [406]

Answer:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

Step-by-step explanation:

Data given and notation

$\bar X_{1}=1.17$ represent the mean for the sample 1 (25 mil film)

$\bar X_{2}=1.04$ represent the mean for the sample 2 (20 mil film)

$s_{1}=0.11$ represent the sample standard deviation for the sample 1

$s_{2}=0.09$ represent the sample standard deviation for the sample 2

$n_{1}=8$ sample size selected for 1

$n_{2}=8$ sample size selected for 2

$\alpha=0.05$ 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 reducing the film thickness increases the mean speed of the film, the system of hypothesis would be:

Null hypothesis: $\mu_{1} \leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

P-value

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

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

4 0

2 years ago

Is it positive or negative or none

Marat540 [252]

Positive because it is increasing

8 0

3 years ago

Read 2 more answers

Nina and Jo both ran an 8 km race. Nina took 55 minutes to run the whole race. Jo started the race 3 minutes later than Nina but

grin007 [14]

Answer:

Step-by-step explanation:

First we figure out how fast Nina can run. If Nina can run 8 km in 55 minutes, then her rate is

$\frac{8km}{55min}=.145\frac{km}{min}$ and we can use that in a d = rt table:

d = r * t

Nina .145

Now we can fill in the distance which is 6 for both, since that is the distance where they met:

d = r * t

Nina 6 = .145

Jo 6 =

Now we go to the info given about the time. If Jo started the race 3 minutes after Nina, that means that Nina is running 3 minutes longer than Jo. Filling in the time info:

d = r * t

Nina 6 = .145 * t + 3

Jo 6 = r * t

As you can see, right now we have 2 unknowns in Jo's row. But we don't have to! We will go to Nina's row where the only unknown is time and solve for t. If d = rt, then

6 = .145(t + 3) and

6 = .145t + .435 and

5.55 = .145t so

t = 38.379 minutes. This means that Jo was running 38.379 minutes when she caught up to Nina (it took Nina 3 minutes longer than that to go 6 km since she was already running for 3 minutes when Jo started the race). If Jo's time is 38.379, we can use that in her row for t and solve for r. If d = rt, then

6 = r(38.379) and

r = .16 km/min

Let's check it without the rounding (rounding takes away from the accuracy). If 6 = .145(t + 3) and Nina's rate not rounded is .145454545 and t = 38.37931034, then, rewriting without rounding:

6 should equal .145454545( 38.37931034 + 3)

6 ?=? .145454545(41.37931034)

6 ?=? 6.0 so

Jo's rate is .16 km/min rounded

6 0

2 years ago