All categories

3 years ago

the mass of a dust particle is 0.000000000735 kilograms. What is the mass written in scientific notation?

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

6 0

0.000000000735 = 7.35 x 10^-10 kg

levacccp [35]3 years ago

4 0

Answer:

Option D - $7.35\times 10^{-10}$

Step-by-step explanation:

Given : The mass of a dust particle is 0.000000000735 kilograms.

To find : What is the mass written in scientific notation?

Solution :

Scientific notation is writing a number into smaller or larger number.

Mostly scientific number is in a number before decimal and then the multiple of 10 to the powers.

So, $0.000000000735=\frac{735}{1000000000000}$

$0.000000000735=7.35\times 10^{-10}$

The mass in scientific notation is $0.000000000735=7.35\times 10^{-10}$ kg.

Therefore, Option D is correct.

You might be interested in

Which expression is equivalent to cube root of 216x27

mariarad [96]

Hello from MrBillDoesMath!

Answer:

Step-by-step explanation:

See attachment

Thank you,

MrB

3 0

3 years ago

A consumer agency is investigating the blowout pressures of Soap Stone tires. A Soap Stone tire is said to blow out when it sepa

Likurg_2 [28]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: Impact force needed for Sap Stone tires to blow out. (foot-pounds)

n= 29, S= 1358 foot-pound

Soap Stone claims that "The tires will blow out at an average pressure of μ= 26000 foot-pounds with a standard deviation of σ= 1020 foot-pounds.

According to the consumer's complaint, the variability of the blown out forces is greater than the value determined by the company.

Then the parameter of interest is the population variance (or population standard deviation) and to test the consumer's complaint you have to conduct a Chi-Square test for σ².

σ²= (1020)²= 1040400 foot-pounds²

H₀: σ² ≤ 1040400

H₁: σ² > 1040400

α: 0.01

II)

$X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}$

$X^2_{H_0}= \frac{(n-1)S^2}{Sigma^2}= \frac{(29-1)*(1358)^2}{1040400} = 49.63$

III)

This test is one-tailed to the right and so is the p-value. This distribution has n-1= 29-1= 28 degrees of freedom, so you can calculate the p-value as:

P(X²₂₈≥49.63)= 1 - P(X²₂₈<49.63)= 1 - 0.99289= 0.00711

⇒ The p-value is less than the significance level so the test is significant at 1%. You can conclude that the population variance of the blowout forces is less than 1040400 foot-pounds², at the same level the population standard deviation of the blow out forces is less than 1020 foot-pounds.

99% CI for the variance. Using the X² statistic you can calculate it as:

$[\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ]$