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Ierofanga [76]
3 years ago
13

Write the expression as a single natural logarithm. 5 In 3

Mathematics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

5ln3=ln(3^5)

Step-by-step explanation:

Given: 5ln(3)

Use rule: alog(b)=log(b^a), aln(b)=ln(b^a) (doesn't matter what the log base is)

Apply rule: ln(3^5)

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Please answer!!!! will make brainliest!!!
hammer [34]

Answer:

B

Step-by-step explanation:

1.

  • 3(x+3) - 2 < 4
  • 3x + 9 -2 < 4
  • 3x < 4-7
  • 3x < - 3
  • x < -1
  • x = (-oo, -1)

2.

  • 1 - x ≤ -1
  • - x ≤ -1 - 1
  • -x ≤ -2
  • x ≥ 2
  • x = [2, +oo)

Correct choice is B.

7 0
3 years ago
Find the equation of a line through (0, 1) with slope m = 2.
Svetllana [295]
That would be : y = 2x + 1
4 0
2 years ago
Find the distance between the two points. (5,-3) (-4,7)
STatiana [176]

Answer:

-9.4

Step-by-step explanation:

8 0
3 years ago
HELP HELP WITH THIS PLZ
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10•10•10=1,000=10^3
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3 years ago
Starting in the 1970s, medical technology allowed babies with very low birth weight (VLBW, less than 1500 grams, about 3.3 pound
cluponka [151]

Answer:

The test statistic value using the VLBW babies as group 1 is z=-2.76±0.01 and the P-value for the test (±0.0001) is 0.0030.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

Then, the null and alternative hypothesis are:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0

The significance level is 0.05.

The sample 1 (VLBW group), of size n1=244 has a proportion of p1=0.77049.

p_1=X_1/n_1=188/244=0.77049

The sample 2 (control group), of size n2=247 has a proportion of p2=0.79757.

p_2=X_2/n_2=197/247=0.79757

The difference between proportions is (p1-p2)=-0.02708.

p_d=p_1-p_2=0.77049-0.79757=-0.02708

The pooled proportion, needed to calculate the standard error, is:

p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{188+197}{244+247}=\dfrac{385}{491}=0.988

The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.988*0.012}{244}+\dfrac{0.988*0.012}{247}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00005}=\sqrt{0.0001}=0.0098

Then, we can calculate the z-statistic as:

z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.02708-0}{0.0098}=\dfrac{-0.02708}{0.0098}=-2.76

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

P-value=P(t

As the P-value (0.0030) is smaller than the significance level (0.05), the effect issignificant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

3 0
3 years ago
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