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jonny [76]
3 years ago
6

PLEASE HELP WITH THIS

Mathematics
1 answer:
Tpy6a [65]3 years ago
6 0

Answer:

the last one

f ^ - 1 * (x) = log5 x

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One quarter of a bread recipe calls for 2/3 cup of bread flour how many cups are needed per recipe
nikklg [1K]

Answer:

2 and 2/3

Step-by-step explanation:

2/3 multiplied by 4

6 0
3 years ago
The coordinates of trapezoid EFGH are E(8, 8), F(8, 10), G(1, 6), and H(4, 1). The image of EFGH under dilation is E'F'G'H'.
VashaNatasha [74]
The coordinates of vertex F' would be (4,5). Since your original coordinates are twice the new coordinates, you'd have to divide the old coordinates by 2 to get the new coordinates of (4,5).
7 0
3 years ago
The 2014 draft picks for the NBA basketball teams have heights that are approximately normally distributed with mean 79.1 inches
ryzh [129]

It is given that the mean is 79.1 inches and standard deviation is 3.0 inches so it follows:

\mu=79.1,\sigma=3

So the value from 75 to 81 is given by:

\begin{gathered} P(75\leq x\leq81)=P(x\leq81)-P(x\leq75) \\ =P(z\leq\frac{81-79.1}{3})-P(z\leq\frac{75-79.1}{3}) \\ =P(z\leq0.63)-P(z\leq-1.36) \end{gathered}

Use the table or a software to get the probability values and substitute to get the probability value as:

The probability is 0.6509.

3 0
1 year ago
If 10000 is invested at an interest rate of 10 per year ,compound semiannually,find the value of the investment after the given
slega [8]

Answer:

Part a) \$17,958.56  

Part b) \$32,251.00  

Part c) \$57,918.16

Step-by-step explanation:

<u><em>The complete question is</em></u>

If $10,000 is invested at an interest rate of 10% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)

a)6 years

b)12 years

c)18 years

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

Part a) 6 years

we have  

t=6\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*6}  

A=10,000(1.05)^{12}  

A=\$17,958.56  

Part b) 12 years

we have  

t=12\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*12}  

A=10,000(1.05)^{24}  

A=\$32,251.00  

Part c) 18 years

we have  

t=18\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*18}  

A=10,000(1.05)^{36}  

A=\$57,918.16

7 0
4 years ago
Starting at sea level, a submarine descended at a constant rate to a depth of −4/5 mile relative to sea level in 3 minutes.
blagie [28]

Answer:

14/5

Step-by-step explanation:

8 0
3 years ago
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