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4 years ago

Ten times a number and nine is 320

1 answer:

6 0

Answer: 10x+9=320 || if you’re looking for the missing number you have to solve for x so move the nine over 10x=320-9, 10x=311, x=31.1

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What is 4/3 plus 3/4

skelet666 [1.2K]

4/3 + 3/4
Make the denominator the same: 12
16/12 + 9/12
= 25/12
Turn that into a mixed number:
= 2 1/25

7 0

3 years ago

Read 2 more answers

What is the set of all solutions to the equation \sqrt{x + 2}= {-x} x+2 =−xsquare root of, x, plus, 2, end square root, equals

NARA [144]

Answer:

$x = 2$ or $x = -1$

Step-by-step explanation:

Given

$\sqrt{x + 2}= {-x}$

Required

Find the possible values fo x

$\sqrt{x + 2}= {-x}$

Square both sides

$( \sqrt{x + 2})^2= ({-x})^2$

$x + 2 = x^2$

Equate to 0

$x^2 - x - 2 = 0$

Expand

$x^2 + x -2x - 2 = 0$

Factorize:

$x(x + 1) - 2(x + 1) = 0$

$(x -2)(x +1) = 0$

Split:

$x - 2 = 0$ or $x + 1 = 0$

$x = 2 + 0$ or $x = 0-1$

$x = 2$ or $x = -1$

3 0

3 years ago

Solve the linear equation 9x-2x+6=4x+3x+11-5

AfilCa [17]

Answer:

Undefined

Step-by-step explanation:

0=0

4 0

3 years ago

Read 2 more answers

Round 43.076 to the nearest hundredth.

AleksAgata [21]

43.08 because if you round and the number in the thousanths is 6 and if the number is 5 and above you round and 4 and below you go down so it is 6 so you round up so it is 43.08 because 7 rounds to 10

8 0

4 years ago

The Institute of Management Accountants (IMA) conducted a survey of senior finance professionals to gauge members’ thoughts on g

DochEvi [55]

Answer:

(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

Step-by-step explanation:

Let X = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.65.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10$

Thus, a Normal approximation to binomial can be applied.

So, $\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)$ .

(1)

Compute the value of $P(0.64$ as follows:

$P(0.64$

$=P(-0.20$

Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 90%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of z for P (Z < z) = 0.95 is

z = 1.65.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73$

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 95%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of z for P (Z < z) = 0.975 is

z = 1.96.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75$

Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

7 0

4 years ago