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viva [34]
3 years ago
5

Jose ha encontrado varios insectos. Hay 10 menos saltamontes que grillos. Hay 5 menos grillos que margaritas. Si Jose ha encontr

ado 5 saltamontes ¿cuantas mariquitas encontró y cuantos grillos? Escribe dos ecuaciones para resolver el problema.
Mathematics
1 answer:
rewona [7]3 years ago
6 0

Answer: There are 5 grasshoppers, 15 crickets and 20 ladybugs

Step-by-step explanation:

The question in english is:

Jose has found several insects. There are 10 less grasshoppers than crickets. There are 5 less crickets than ladybugs. If Jose has found 5 grasshoppers, how many ladybugs did he find and how many crickets? Write two equations to solve the problem.

Let's tag grasshoppers with g, crickets with c and ladybugs with c.

So, we are tolde there are 10 less grasshoppers than crickets:

g=c-10 (1) We have the first equation

Then, we are told there are 5 less crickets than ladybugs:

c=l-5 (2) This is the second equation

If g=5 and we substitute this in both equations we will have:

5=c-10 (3)

Isolating c:

c=15 (4) There are 15 crickets

Substituting (4) in (2):

15=l-5 (5)

Isolating l:

l=20 There are 20 ladybugs

Therefore:

There are 15 crickets and 20 ladybugs

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Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

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Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

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The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

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