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2 years ago

HELP FAST!!! MULTIPLE CHOISE!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Dvinal [7]2 years ago

6 0

Answer: C. 1462

Step-by-step explanation: It is 1462 Kilograms, why? When rounding the 6 in the tens place rounds the 4 up. Which leads to it being 1500 Kilograms. Hope this clears up your problem! :P

liq [111]2 years ago

5 0

letter C

1462 rounds up when looking at the tens place to 1500

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The lengths of a rectangle have been measured to the nearest tenth of a centimetre they are 87.3cm and 51.8cm what is the upper

vagabundo [1.1K]

Answer: Area (upper bound) = 4527.7056 cm²

Perimeter (lower bound) = 278 cm

Step-by-step explanation:

The length and width of the rectangle have been ROUNDED to the nearest tenth. Let's calculate what their actual measurements could be:

LENGTH: rounded to 87.3, actual is between 87.25 and 87.34

87.25 is the lowest number it could be that would round it UP to 87.3

87.34 is the highest number it could be that would round DOWN to 87.3

WIDTH: rounded to 51.8, actual is between 51.75 and 51.84

51.75 is the lowest number it could be that would round it UP to 51.8

51.84 is the highest number it could be that would round DOWN to 51.8

To find the Area of the upper bound, multiply the highest possible length and the highest possible width:

A = 87.34 × 51.84 = 4527.7056

To find the Perimeter of the lower bound, calculate the perimeter using the lowest possible length and the lowest possible width:

P = 2(87.25 + 51.75) = 278

8 0

2 years ago

Read 2 more answers

Can some one please help me????

Juli2301 [7.4K]

$-\frac{3}{2}w-\frac{7}{8}=\frac{1}{2}|\cdot8\\
-12w-7=4\\
-12w=11\\
w=-\frac{11}{12}
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2 years ago

What is 29/12 as a decimal

ivolga24 [154]

Answer:

2.4166666666 (6 forever)

Step-by-step explanation:

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2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

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

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

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

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

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