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2 years ago

One and three fourths subtracted from seven tenths

Mathematics

1 answer:

Brut [27]2 years ago

3 0

First convert both numbers into improper fractions. Then find the least (not greatest, photo is incorrect) common denominator and multiple by how much depending on your fraction. After that subtract and simplify needed.

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Elevator 1 in a building moved from ground position to a final position of +16 feet. Elevator 2 in the same building moved from

lilavasa [31]

928282827282828382082820ks

8 0

2 years ago

Suppose you have a pocketful of change. You have some pennies (p) and some quarters (q). One expression could be used to describ

faltersainse [42]

Answer:

1.57

Step-by-step explanation:

So we have two equations regarding the number of quarters q and the number of pennies p:

$p+q$

Which represents the total amount of coins and

$0.01p+0.25q$

Which represents the total amount of money in dollars.

So we are asked to evaluation each expression for the situation in which we have 6 quarters and 7 pennies. Thus, plug 6 in for q and 7 in for p:

$p+q\\(7)+(6)=13$

This tells us that we have 16 coins in total.

$0.01(7)+0.25(6)\\=0.07+1.50\\=1.57$

This tells us that we have a total amount of $1.57.

5 0

2 years ago

Read 2 more answers

Gravel is being dumped from a conveyor belt at a rate of 15 ft3/min, and its coarseness is such that it forms a pile in the shap

gayaneshka [121]

Answer:

0.13 ft/min

Step-by-step explanation:

We are given that

$\frac{dV}{dt}=15ft^3/min$

We have to find the increasing rate of change of height of pile when the pile is 12 ft high.

Let d be the diameter of pile

Height of pile=h

d=h

Radius of pile,r= $\frac{d}{2}=\frac{h}{2}$

Volume of pile= $\frac{1}{3}\pi r^2 h=\frac{1}{12}\pi h^3$

$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$

h=12 ft

Substitute the values

$15=\frac{1}{4}\pi(12)^2\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{15\times 4}{\pi(12)^2}$

$\frac{dh}{dt}=0.13ft/min$

7 0

2 years ago

Sam invests $6000 in two different accounts. The first account paid 12 %, the second account paid 7 % in interest. At the end of

Lunna [17]

Answer:

The amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600

Step-by-step explanation:

Let x be the amount in the account at 12% interest

So, 6000-x is the amount in the second account at 7% interest

$SI = \frac{P \times T \times R}{100}$

First account: $SI=\frac{x \times 1 \times 12}{100}$

Second account : $SI =\frac{(6000-x) \times 1 \times 7}{100}$

We are given that At the end of the first year he had earned $590 in interest.

So, $\frac{x \times 1 \times 12}{100}+\frac{(6000-x) \times 1 \times 7}{100}=590\\x=3400$

So,the amount in the account at 12% interest is $3400

The amount in the second account at 7% interest =6000-x=6000-3400=2600

Hence the amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600

8 0

2 years ago

Iq scores are normally distributed with a mean of 100 and a standard deviation of 15.

bearhunter [10]

Mean is $\mu=100$ and a standard deviation is $\sigma =15$ , then the variable $X\sim N(100,15^2)$ .

Use substitution

$Z=\dfrac{X-\mu}{\sigma},\\ \\ Z=\dfrac{X-100}{15}$ .

This substitution gives you that $Z\sim N(0,1)$ .

a. For X=130, $Z=\dfrac{130-100}{15}=2$ and $Pr(X>130)=Pr(Z>2) =0.9772$ (the decimal value is taken from the Standard Normal Distribution Table).

b. For X=90, $Z=\dfrac{90-100}{15}=-\dfrac{2}{3}$ and for X=110, $Z=\dfrac{110-100}{15}=\dfrac{2}{3}$ . Then $Pr(90$ (the decimal value is taken from the Standard Normal Distribution Table).

6 0

2 years ago