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3 years ago

Remove the parentheses from the expression (–2) – (–3). A. 2 – 3 B. –2 + 3 C. 2 + 3 D. –2 – 3

Mathematics

1 answer:

lukranit [14]3 years ago

5 0

Answer:

Step-by-step explanation:

The negative of a negative is a positive, so B is your choice. Think of it anothe way, if it helps. Imagine that the expression is this:

(-2) + (-1)(-3) since every subtraction problem can be rewritten as an addition problem.

We would first, by orders of operation, multiply the -1 and the -3 getting 3. Then the problem would simplify down to -2 + 3

Just another idea to ponder...

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Which ones are rational and which are irrational?

In-s [12.5K]

Answer:

irrational, irrational, irrational, rational, rational

Step-by-step explanation:

3 0

2 years ago

5 × 10 to the power of 22 is how many times greater than 1 × 10 to the power of 21

alekssr [168]

Answer:

50 times greater

Step-by-step explanation:

If there are two numbers A and B, where A is greater than B,

If we want to find how many times greater A is to B, we simply divide, so that would be

A/B

Now, we have

$5*10^{22}$

and

$1*10^{21}$

We simply need to divide to find how much greater is the first number than the second number. Remember dividing numbers in scientific notation:

$\frac{a*10^b}{c*10^d}\\=(\frac{a}{c})*10^{b-d}$

Thus, we have:

$\frac{5*10^{22}}{1*10^{21}}\\=(\frac{5}{1})*10^{22-21}\\=5*10^{1}\\=5*10\\=50$

So the first number is 50 times greater than the second

4 0

3 years ago

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

2 years ago

Finn bought 2 packs of stickers. A friend gave him 4 more stickers. Now he has 24 stickers in all. How many stickers were in eac

Murljashka [212]

2x+4=24 -4 -4 ------------- 2x=20 ÷2 ÷2 ---------- X=10 So 10 stickers in each pack

6 0

3 years ago

Read 2 more answers

The output from a statistical computer program indicates that the mean and standard deviation of a data set consisting of 200 me

lesya [120]

Answer:

The limit that 97.5% of the data points will be above is $912.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 1500, \sigma = 300$