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3 years ago

50 POINTS + BRAINLIEST

Mathematics

1 answer:

yulyashka [42]3 years ago

8 0

Answer:

522 $m^2$

Step-by-step explanation:

To find the surface area, we can find the area of each shape in the net.

There are 2 identical triangles and 3 different rectangles. Then we can add all of these areas together

$2(\frac{1}{2} (9)(14))=126\\\\(10)(14)=140\\\\(10)(16.6)=166\\\\(9)(10)=90\\\\126+140+166+90=522$

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Could someone help me understand how to solve this?

monitta

Answer:

Amount in 4% account: $550

Step-by-step explanation:

Use simple interest formula $I=P\cdot r\cdot t,$ where

I = interest

P = principal

r = rate (as decimal)

t = time.

<u>4% account:</u>

$P_1=x\\ \\r_1=0.04\\ \\t_1=1,$

then

$I_1=x\cdot 0.04\cdot 1=0.04x$

<u>5% account:</u>

$P_2=1,500-x\\ \\r_2=0.05\\ \\t_2=1,$

then

$I_2=(1,500-x)\cdot 0.05\cdot 1=0.05(1,500-x)$

You have earned $69.50 in total, so

$I_1+I_2=69.50\\ \\0.04x+0.05(1,500-x)=69.50\\ \\4x+5(1,500-x)=6,950\\ \\4x+7,500-5x=6,950\\ \\4x-5x=6,950-7,500\\ \\-x=-550\\ \\x=550\\ \\1,500-x=1,500-550=950$

Amount in 4% account: $550

Amount in 5% account: $950

8 0

3 years ago

Chef Morgan’s restaurant has 24 tables. Fifteen of the tables each have one flower in a vase. The remaining tables have 5 flower

AveGali [126]

I think there is 45 flowers in total.

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3 years ago

Read 2 more answers

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$