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3 years ago

Using pemdas get the answer for -6-5(4-6)

Mathematics

2 answers:

Veronika [31]3 years ago

7 0

Answer:

Step-by-step explanation:

-6-5(4-6)

PEMDAS stands for parenthesis, exponent, multiply, divide, add, subtract. Let's follow those rules!

First, we'll do the parenthesis.

-6-20+30

Now we can just add and subtract everything together.

-6+10

MA_775_DIABLO [31]3 years ago

4 0

Answer: 4

Step-by-step explanation:

PEMDAS is an order-of-operations strategy that helps you remember the correct order.

Parentheses, exponents, multiplication/division, addition/subtraction.

Your equation is:

$-6-5*(4-6)$

First, deal with parentheses.

$-6-5*-2$

Next, you can multiply -5 and -2 to get 10.

$-6 + 10$

Your answer is 4.

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The price have changed from $120 to $75. By how many

Kitty [74]

Answer:

35.7%

Step-by-step explanation:

The change is 45 dollars. (120-75)

The percent of change is the dollars divided by the original price.

45/120=0.357

Or 35.7%

3 0

3 years ago

Read 2 more answers

Factor the following polynomial.<br><br> ( Correct and best explained answer gets brainliest. )

mamaluj [8]

Glad you asked this question! Over the years of working with my professor we learned that if phs= 5 (constant) times the j to the power of 2 you will get the formula or polynomialic equation. It quite fascinating and due to using his equation the answer in fact is "c". Cheers

4 0

3 years ago

What is the 8th term of this geometric sequence 6,48,384,3072

ELEN [110]

Common ratio can be found by dividing the 2nd term by the first
r = 48/6
r = 8

an = a1 * r^(n-1)
n = term to find = 8
a1 = first number = 6
r = common ratio = 8
now we sub
a(8) = 6 * 8^(8-1)
a(8) = 6 * 8^7
a(8) = 6 * 2097152
a(8) = 12582912 <==

3 0

3 years ago

Use the standard normal distribution or the t-distribution to construct a 99% confidence interval for the population mean. Jus

vichka [17]

Answer:

(25.732,30.868)

Step-by-step explanation:

Given that in a random sample of 42 people, the mean body mass index (BMI) was 28.3 and the standard deviation was 6.09.

Since only sample std deviation is known we can use only t distribution

Std error = $\frac{s}{\sqrt{n} } =\frac{6.09}{\sqrt{42} } \\=0.9397$

$df = 42-1 =41$

t critical for 99% two tailed $= 2.733$

Margin of error $= 2.733*0.9397=2.568$

Confidence interval lower bound = $28.3-2.568=25.732$

Upper bound = $28.3+2.568=30.868$

3 0

3 years ago

6-10 divide, another onee thank you!!

eduard

Answers:

10.) $\displaystyle \pm{5}$

9.) $\displaystyle 1\frac{1}{2}$

8.) $\displaystyle \pm{1\frac{1}{2}}$

7.) $\displaystyle \pm{1\frac{1}{2}}$

6.) $\displaystyle \pm{\frac{1}{2}}$

Step-by-step explanations:

10.) $\displaystyle \frac{\sqrt{200}}{\sqrt{8}} \hookrightarrow \sqrt{25} \hookrightarrow \frac{\pm{10\sqrt{2}}}{\pm{2\sqrt{2}}} \\ \\ \boxed{\pm{5}}$

9.) $\displaystyle \frac{\sqrt[3]{135}}{\sqrt[3]{40}} \hookrightarrow \sqrt[3]{3\frac{3}{8}} \hookrightarrow \frac{3\sqrt[3]{5}}{2\sqrt[3]{5}} \\ \\ \boxed{1\frac{1}{2}}$

8.) $\displaystyle \frac{\sqrt[4]{162}}{\sqrt[4]{32}} \hookrightarrow \sqrt[4]{5\frac{1}{16}} \hookrightarrow \frac{\pm{3\sqrt[4]{2}}}{\pm{2\sqrt[4]{2}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}$

7.) $\displaystyle \frac{\sqrt{63}}{\sqrt{28}} \hookrightarrow \sqrt{2\frac{1}{4}} \hookrightarrow \frac{\pm{3\sqrt{7}}}{\pm{2\sqrt{7}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}$

6.) $\displaystyle \frac{\sqrt{12}}{\sqrt{48}} \hookrightarrow \sqrt{\frac{1}{4}} \hookrightarrow \frac{\pm{2\sqrt{3}}}{\pm{4\sqrt{3}}} \\ \\ \boxed{\pm{\frac{1}{2}}}$

I am joyous to assist you at any time.

5 0

2 years ago