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4 years ago

Homework

Mathematics

1 answer:

jenyasd209 [6]4 years ago

6 0

Answer:

6,9 hope this is it

Step-by-step explanation:

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COS Find the quotient when the polynomial expression is divided by a binomial Find the quotient ANSWER SOLUTION CHOICES X2+11x+

ryzh [129]

Answer:

x=−1 or x=−9

Step-by-step explanation:

x2+11x+10=x+1

Step 1: Subtract x+1 from both sides.

x2+11x+10−(x+1)=x+1−(x+1)

x2+10x+9=0

Step 2: Factor left side of equation.

(x+1)(x+9)=0

Step 3: Set factors equal to 0.

x+1=0 or x+9=0

x=−1 or x=−9

8 0

3 years ago

Caroline is making some table decorations.

MrMuchimi

Answer:

Caroline buys 3 packs of candles and 5 packs of holders.

Step-by-step explanation:

1.) First, find multiples of 30 by adding 30 to itself. For example, 30, 60, 90.

2.) Second, find multiples of pakcs of holders until it matches the multiples of candle packs. 18, 36, 54, 72, 90.

3.) then you find that Caroline buys 3 packs of candles and 5 packs of candle holders so she has the same amount of both.

4 0

3 years ago

Which is equivalent to 64 Superscript one-fourth?

Schach [20]

Answer:

2 RootIndex 4 StartRoot 4 EndRoot

Step-by-step explanation:

we have

$64^{\frac{1}{4}}$

Decompose the number 64 in prime factors

$64=2^{6}=2^{4}2^{2}$

substitute

$64^{\frac{1}{4}}=(2^{4}2^{2})^{\frac{1}{4}}=2^{\frac{4}{4}}2^{\frac{2}{4}}=2\sqrt[4]{4}$

7 0

3 years ago

Read 2 more answers

Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i

OLEGan [10]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

$H_0:\mu=18\\\\H_a:\mu\neq18$

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size : n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : $\overline{x}=18.8$

Standard deviation : $\sigma=2.8$

Test statistic for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02$

The p-value= $2P(z>2.02)=0.0433834$

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0

3 years ago

Why are prefixes used in naming covalent compounds

Nataly_w [17]

Answer:

the atoms can have different numbers of valence electrons

6 0

2 years ago