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3 years ago

5

Enrique is measuring the mass of a flower petal. Which unit of measurement should he use to eliminate the need to write the valu

e in scientific notation? grams. centimeters. liters. kilograms.

2 answers:

GuDViN [60]3 years ago

6 0

Answer:

do you have a pic of it.

Step-by-step explanation:

alukav5142 [94]3 years ago

5 0

Answer:

Grams

Step-by-step explanation:

Edg. 2020 :D

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1) If the designer measures the pond and determines that the length of the pond across is 400

ozzi

Answer: 200

Step-by-step explanation: to get the radius you divided by 2. bc there is only 2 sides , have a great day.

3 0

3 years ago

A rectangle has vertices E (-4,8), F(2,8), G(2,-2) and H (-4,-2). The rectangle is dilated with the origin as the center of dila

bezimeni [28]

Answer:

3.5

Step-by-step explanation:

The

3 0

3 years ago

Need help asappp special right triangle

Korvikt [17]

Answer:

your answer is 30

Step-by-step explanation:

p to r is 15

p to q is 15 beacuse these lines are the same

and 15 plus 15 is 30

3 0

3 years ago

The altitude of a triangle is increasing at a rate of 1.5 1.5 centimeters/minute while the area of the triangle is increasing at

seraphim [82]

Answer:

The base reduces at 3.75cm/min

Step-by-step explanation:

Given

Let

$h \to altitude$

$b \to base$

$A \to Area$

So:

$\frac{dh}{dt} = 1.5cm/min$

$\frac{dA}{dt} = 1.5^2cm/min$

The area of a triangle is:

$A = \frac{1}{2}bh$

Calculate b when $A =88cm^2; h =8cm$

$A = \frac{1}{2}bh$

$88=\frac{1}{2} * b * 8$

$88 =b * 4$

Solve for b

$b = 88/4$

$b = 22$

We have:

$A = \frac{1}{2}bh$

Differentiate with respect to time

$\frac{dA}{dt} =\frac{1}{2}(h\frac{db}{dt} + b\frac{dh}{dt})$

Substitute the following values in the above equation

$\frac{dh}{dt} = 1.5cm/min$ $\frac{dA}{dt} = 1.5^2cm/min$ $b = 22$ $h = 8$

$1.5 = \frac{1}{2}(8 * \frac{db}{dt} + 22 * 1.5)$

Multiply both sides by 2

$3 = 8 * \frac{db}{dt} + 22 * 1.5$

$3 = 8 * \frac{db}{dt} + 33$

Collect like terms

$8 * \frac{db}{dt} = 3 -33$

$8 * \frac{db}{dt} = -30$

Divide both sides by 8

$\frac{db}{dt} = -\frac{30}{8}$

$\frac{db}{dt} = -3.75$

4 0

3 years ago

In a standard Normal distribution, if the area to the left of a z-score is about 0.3500, what is the approximate z-score? Dr

wlad13 [49]

Answer:

z-score=0.385

(See attached picture)

Step-by-step explanation:

The procedure to find the z-score will depend on the resources we have available. I have a table with the area between the mean and the value we wish to normalize, so the very first thing we need to do is precisely find this area we need to analyze.

Everything to the left of thte mean will represent 50% of the data, so we start by subtracting:

50%-35%=15%

so we need to look in the table for the value 0.15.

In my table I can see that for an area of 0.15, the z-score will be between 0.38 (z-score of 0.1480) and 0.39 (z-score of 0.1517).

By doing some interpolation, you can determine a more accurate value of the z-score to be 0.385.

4 0

4 years ago