Express 2.3×(10)^-5 in standard form

If 4 apples have a mass of 1 kg. About how many grams is each apple?

igomit [66]

Answer:

0.25kg

Step-by-step explanation:

1kg/4(total divided by apples)=0.25kg

3 0

4 years ago

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Adding and subtracting intergers. FREE BRAINLIST AND POINTS!. WRONG OR SCAM ASNWERS WILL BE REPORTED AND DEALED WITH. RIGHT ANSW

s344n2d4d5 [400]

Answer:

1. 21

2. -22

3. -19

4. -17

5. -5

6. 11

7. 7

8. -9

9. -6

10. -20

11. -1

12. 20

13. -21

Step-by-step explanation:

5 0

3 years ago

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Suppose that the probability density function ofthe length of computer cables is f (x) = 0.1 from 1200 to1210 millimeters.a. Det

Ann [662]

Notice that $f(x)=0.1$ for $1200\le x\le1210$ implies that $f(x)=0$ elsewhere, since

$\displaystyle\int_{1200}^{1210}f(x)\,\mathrm dx=1$

where $X=x$ is a random variable representing cable lengths according to the PDF $f(x)$ .

a. By definition of expectation, the mean is

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f(x)\,\mathrm dx=0.1\int_{1200}^{1210}x\,\mathrm dx=1205$

The variance is

$\operatorname{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f(x)\,\mathrm dx=0.1\int_{1200}^{1210}x^2\,\mathrm dx=\frac{4,356,100}3$

so that the variance is $\frac{25}3$ , making the standard deviation $\frac5{\sqrt3}$ .

b. The proportion of cables within specs is

$P(1195$

8 0

4 years ago

What is the area of that figure??

pochemuha

Answer:

It's 51 square inches. You can multiply 12x8 and subtract 9x5 from that.

4 0

3 years ago

When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted

hodyreva [135]

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ = 0.

Hₐ: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ ≠ 0.

Assume that the significance level of the test is, α = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

Compute the pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16$

The critical value of t is:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984$

*Use a t-table.

Compute the 95% confidence interval for mean difference as follows:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

$=(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}$

$=-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)$

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.

7 0

4 years ago