Help a brother out!!

What is the best metric unit to use for the mass of a grasshopper

mixas84 [53]

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷The best metric unit to use for the mass of a grasshopper is grams, because a grasshopper is very light and a gram is small.

One gram is about the weight of a paper clip or a butterfly.

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

TROLLER

6 0

3 years ago

Read 2 more answers

Find the area of the shape below

maks197457 [2]

Answer:

204 ft^2

Step-by-step explanation:

triangle on the left:

A= l x w x 1/2

A = 12 x 5 x 1/2

A = 30 ft^2

triangle on the right:

Same thing

A = 30 ft^2

square in the middle:

A = l x w

A = 12 x 12

A = 144 ft^2

Add them up:

30 + 30 + 144 = 204 ft^2

3 0

3 years ago

A city had population 67,255 on january 1, 2000, and its population has been increasing by 2935 people each year since then. A l

irina1246 [14]

We are given: On january 1, 2000 initial population = 67,255.

Number of people increase each year = 2935 people.

Therefore, 67,255 would be fix value and 2935 is the rate at which population increase.

Let us assume there would be t number of years after year 2000 and population P after t years is taken by function P(t).

So, we can setup an equation as

Total population after t years = Number of t years * rate of increase of population + fix given population.

In terms of function it can be written as

P(t) = t * 2935 + 67255.

Therefore, final function would be

P(t) = 2935t +67255.

So, the correct option is d.P(t) = 67255 + 2935t.

4 0

3 years ago

Read 2 more answers

In the given figure △ABC ≅△DEC. Which of the following relations can be proven using CPCTC ?

Serhud [2]

Option B:

$\overline{A B}=\overline{D E}$

Solution:

In the given figure $\triangle A B C \cong \triangle D E C$ .

If two triangles are similar, then their corresponding sides and angles are equal.

By CPCTC, in $\triangle A B C \ \text{and}\ \triangle D E C$ ,

$\overline{AB }=\overline{DE}$ – – – – (1)

$\overline{B C}=\overline{EC}$ – – – – (2)

$\overline{ CA}=\overline{CD}$ – – – – (3)

$\angle ACB=\angle DCE$ – – – – (4)

$\angle ABC=\angle DEC$ – – – – (5)

$\angle BAC=\angle EDC$ – – – – (6)

Option A: $\overline{B C}=\overline{D C}$

By CPCTC proved in equation (2) $\overline{B C}=\overline{EC}$ .

Therefore $\overline{B C}\neq \overline{D C}$ . Option A is false.

Option B: $\overline{A B}=\overline{D E}$

By CPCTC proved in equation (1) $\overline{AB }=\overline{DE}$ .

Therefore Option B is true.

Option C: $\angle A C B=\angle D E C$

By CPCTC proved in equation (4) $\angle ACB=\angle DCE$ .

Therefore $\angle A C B\neq \angle D E C$ . Option C is false.

Option D: $\angle A B C=\angle E D C$

By CPCTC proved in equation (5) $\angle ABC=\angle DEC$ .

Therefore $\angle A B C\neq \angle E D C$ . Option D is false.

Hence Option B is the correct answer.

$\Rightarrow\overline{A B}=\overline{D E}$

5 0

3 years ago

Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 a

krok68 [10]

Answer:

Jessie scored higher than Reagan.

Step-by-step explanation:

We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.

Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.

For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.

1. Finding z-score for Reagan;

Let X = distribution of SAT scores

SO, X ~ Normal( $\mu=1000, \sigma^{2}=100^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 1000

$\sigma$ = standard deviation = 100

Now, Reagan scored 1140 on the SAT, that is;

z-score = $\frac{1140-1000}{100}$ = 1.4

2. Finding z-score for Jessie;

Let X = distribution of ACT scores

SO, X ~ Normal( $\mu=17, \sigma^{2}=5^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 17

$\sigma$ = standard deviation = 5

Now, Jessie scored 30 on the ACT, that is;

z-score = $\frac{30-17}{5}$ = 2.6

This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.

6 0

3 years ago