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3 years ago

Need answers within 10 mins help

Mathematics

1 answer:

Irina-Kira [14]3 years ago

5 0

Answer:

Step-by-step explanation:

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Expand the logarithmic expression.

lapo4ka [179]

The answer for the exercise shown above is the option B, which is:
 B. log5(7)+5log5(a)
The explanation of the problem is shown below:
1. You have the following expression given in the problem above:
log5(7)(a^5)
2. To expand it, you must use the logaritms properties, as following:
 log5(7)(a^5)
log5(7)+log5(a^5)
log5(7)+5log5(a)
3. Therefore, as you can see, the answer is the option B.

8 0

4 years ago

Which point satisfies both ƒ(x) = 2x and g(x) = 3x? (0,1) (0,-1) (1,0) (-1, 0)

ycow [4]

Answer:

(0,1)

Step-by-step explanation:

we are given functions as

$f(x)=2^x$

$g(x)=3^x$

we can verify each options

option-A:

we can plug x=0 and check whether we get y=1

$f(0)=2^0=1$

$g(0)=3^0=1$

so, this is TRUE

option-B:

we can plug x=0 and check whether we get y=-1

$f(0)=2^0=1$

$g(0)=3^0=1$

so, this is FALSE

option-C:

we can plug x=1 and check whether we get y=0

$f(1)=2^1=2$

$g(1)=3^1=3$

so, this is FALSE

option-D:

we can plug x=-1 and check whether we get y=0

$f(-1)=2^{-1}$

$g(-1)=3^{-1}$

so, this is FALSE

5 0

3 years ago

Subtract the quotient of 15 divided by 3 from the sum of 33 and 7

iragen [17]

Answer:

(33+7)-(15/3) = 35

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Con un descuento de 30% pero con un adicional de 16% cuánto pagará en realidad por el televisor

OleMash [197]

Answer:

my boomboom

Step-by-step explanation:

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3 years ago

A genetics experiment involves a population of fruit flies consisting of 1 male named Bart and 3 females named Charlene, Diana,

kenny6666 [7]

Answer:

(b) $\frac{1}{2}$

Step-by-step explanation:

1 male named Bart (b)

3 females named Charlene(c), Diana(d), and Erin(e).

Since there is replacement, the possible samples are:

bb, bc, be, bd, cb, cc, cd, ce, db, dc, dd, de, eb, ec, ed, and ee.

Total Number of pairs = 16

Event of picking 2 males:bb

Event of 1 male:bc,cb,bd,db,be,eb

Event of picking 0 males:

cc, cd, ce, dc, dd, de, ec, ed, and ee.

$\begin{equation*} \begin{matrix}Proportion of males & Probability \\0 & 9/16 \\1 & 6/16 \\2 & 1/16 \end{matrix} \end{equation*}$

b. The mean of the sampling distribution

$\mu = (\frac{9}{16} X0 ) +(\frac{6}{16} X 1)+(\frac{1}{16} X 2) = \frac{8}{16} = \frac{1}{2}$

c. No; the proportion of males is $\frac{1}{4}$ while the mean is $\frac{1}{2}$ .

B. No, the sample mean is not equal to the population proportion of males. These values are not always equal, because proportion is an unbiased estimator.

7 0

3 years ago