Daily highs in Sacramento for the past week were:

Asap

vagabundo [1.1K]

Answer:

The total surface are is 252 sq. units.

Step-by-step explanation:

The right triangular sides have legs 8 units and 6 units and hypotenuse 10 units.

So, their area will be $2(\frac{1}{2} \times 8 \times 6) = 48$ sq. units.

Now, the length of the prism is 8.5 units.

So, the area along the length will be = (8.5 × 6 + 8.5 × 8 + 8.5 × 10) = 204 sq. units.

Therefore, the total surface area is (48 + 204) = 252 sq. units. (Answer)

7 0

3 years ago

(ruu ule

boyakko [2]

Answer:

y-3 = 2x

2x-y+3=0 we have

slope

m =- coefficient of x/coefficient of y=-2/-3=2/3 is a slope..

8 0

3 years ago

Guys I need your help!! Jake makes 9 loaves of olive bread. He uses 30 grams of olives in each loaf. He started with 1 kilogram

Harrizon [31]

Answer:

730 g of olives

Step-by-step explanation:

First, convert 1 kg of olives to 1,000 g of olives.

Since he uses 30 grams per loaf, and makes 9 loaves, multiply 30x9, which equals 270g. This means he used 270 g of olives.

To find how many he has left, you subtract the amount he used from the amount he started with:

1,000 - 270 = 730 g

5 0

3 years ago

Read 2 more answers

Sai has a garden on the side of his house that will be enclosed with 196 ft. of fencing. The

Anna35 [415]

Answer:

Do I have to also use what is classified as 'proper english'(Grammatically correct and suitable) for the answer I write

Step-by-step explanation:

6 0

2 years ago

Bodytemperatures ofhealthy koalasarenormally distributed with a mean of 35.6°C and a standard deviation of 1.3°C. a.What is the

Licemer1 [7]

Answer:

a

$P(X < 35) = 32.3 \%$

b

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

c

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 35.6^oC$

The standard deviation is $s = 1.3^oC$

The sample size is n = 30

Generally the probability that a health koala has a body temperature less than 35.0°C is mathematically represented as

$P(X < 35) = P(\frac{X - \mu }{s} < \frac{35 - 35.6}{1.3} )$

Here $(\frac{X - \mu }{s} = Z (The \ standardized \ value \ of \ X )$

So

$P(X < 35) = P(Z < -0.46)$

From the z-table P(Z < -0.46) = 0.323

So

$P(X < 35) = 0.323$

Converting to percentage

$P(X < 35) = 0.323 * 100$

$P(X < 35) = 32.3 \%$

considering question b

The sample mean is $\= x = 35$

Generally the standard error of the mean is mathematically represented as

$\sigma_{\= x} = \frac{s}{\sqrt{n} }$

=> $\sigma_{\= x} = \frac{1.3}{\sqrt{30} }$

=> $\sigma_{\= x} = 0.2373$

Generally the probability of the mean body temperature of koalas being less than 35.0°C is mathematically represented as

$P(\= X < 35) = P(\frac{\= X - \mu }{\sigma_{\= x }} < \frac{35 -35.6}{0.2373 } )$

$P(\= X < 35) = P(Z< -2.53 )$

From the z-table we have that

$P(Z< -2.53 ) = 0.006$

So

$P(\= X < 35) = 0.006 /tex] Converting to percentage [tex]P(\= X < 35) = 0.006 * 100$

$P(\= X < 35) = 0.6 \%$

Here the probability of koalas mean temperature being less than 35 °C is very small hence the koalas are not healthy

A potential confounding variable for this study is the population of the koalas because in the first question the population was not taken into account and the probability was $P(X < 35) = 32.3 \%$ but when the population was taken into account (i.e n = 30) the probability became

$P(\= X < 35) = 0.6 \%$

4 0

4 years ago