All categories

2 years ago

Need help due in 1 min!! Simplify!!

Mathematics

2 answers:

s2008m [1.1K]2 years ago

5 0

Answer:

3^9

Step-by-step explanation:

5+4=9

Katen [24]2 years ago

4 0

Answer:

C. 3^9

Step-by-step explanation:

You would simply just add the exponents.

You might be interested in

he nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the

MrRissso [65]

Answer:

The 99% confidence interval for the mean sodium content in Oriental Spice Sauce is between 454.67 milligrams and 1722.61 milligrams

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 40 - 1 = 39

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 39 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.7079

The margin of error is:

M = T*s = 2.7079*234.12 = 633.97

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1088.64 - 633.97 = 454.67 milligrams

The upper end of the interval is the sample mean added to M. So it is 1088.64 + 633.97 = 1722.61 milligrams

The 99% confidence interval for the mean sodium content in Oriental Spice Sauce is between 454.67 milligrams and 1722.61 milligrams

8 0

3 years ago

Randy and Susie Moser are newlyweds. Randy drives their car to his job, twenty-seven miles away from home, each day. If Randy an

Andrews [41]

The answer is $834.10

3 0

3 years ago

Read 2 more answers

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means: