All categories

3 years ago

Evaluate.

Mathematics

1 answer:

Daniel [21]3 years ago

3 0

Answer:

d) 192

Step-by-step explanation:

Your calculator can do this for you.

Or, you can simplify to

$12\cdot\dfrac{4^{-2}}{4^{-4}}=12(4^{-2+4})=12(4^2)=192$

You might be interested in

Given PQRS and ZS = 40°, what is the measure of ZR? S R 40°

morpeh [17]

it is 140° Bc angle S and Z should add up to 180

7 0

3 years ago

Read 2 more answers

40 percent of 112.5%

lord [1]

First 112% is 1.12 divide it by 100 to get .0112 then mutiply it by 40 to get .448 so .448 is the answer. Hope this helps.

7 0

3 years ago

Identify a pair of exterior angles. 1 and 5 8 and 4 2 and 5 1 and 8

Alenkasestr [34]

Answer:

1 and 8

Step-by-step explanation:

Extierior think of it as interior is the inside they both start with in and then extierior is the outside

7 0

3 years ago

I Need HelpAsap please I will mark you brainlist

Arada [10]

Answer:

a. 72 inches

b. 21.3 inches

Step-by-step explanation:

12:27

a. 32 inches in height

32/x = 12/27

Set these fractions equal to each other

32= 12

12 times 2.6(repeating decimal) equals 32

So to find the width do 2.666... time 27 and you get 72 inches.

---------------------------------------------------------------------------

b. 48 inches in width

Again set the fractions equal to each other.

x/48 = 12/27

48 divided by 27 equals 1.77777....

To get height, do 1.77777.... times 12 and you get 21.3333333

But since it is to the nearest tenth, you get 21.3 inches

Hope that helps!

5 0

3 years ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 421 gram setting. It is

Anastaziya [24]

Answer:

The p-value is approximately 0.0508

Step-by-step explanation:

The given parameters are;

The expected mass of banana chips bag filled by the machine, μ = 421 gams

The number of bags in the sample of bags filled by the machine, n = 26 bags

The mean mass of the bags in the sample, $\overline x$ = 413 grams

The standard deviation of the mass in the bags, s = 24 grams

The level of significance, α = 0.02

The null hypothesis, H₀; μ = 421 grams

The alternative hypothesis, Hₐ; μ < 421 grams

The test statistic is given as follows;

$t=\dfrac{\bar{x}-\mu }{\dfrac{s}{\sqrt{n}}}$

We get;

$t=\dfrac{413-421 }{\dfrac{24}{\sqrt{26}}} \approx -1.6996731712$

The t-value = -1.6996731712

The degrees of freedom df = n - 1 = 26 - 1 = 25

From an online source, we get the critical-t = 2.166587

The p-value is obtained using an online source as p(t ≤ -1.6996731712) = 0.0508

From the p-value table, we get

0.05 < p < 0.1

3 0

2 years ago