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sammy [17]
3 years ago
9

The low temperature in one city was -4°F. The low temperature in another city was 8°F. Write an inequality to compare the temper

atures.
Mathematics
1 answer:
Vikki [24]3 years ago
8 0

-4 < 8. Negatives are lower than positives so the inequality should be negative is less than 8
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Rick needs to be at school at 8:15 A.M. It takes him 20 minutes to walk to school.At what time dose he need to leave to get to s
Rom4ik [11]
Rick needs to leave at 7:55am to get to school on time.

7:55am is 20 minutes before 8:15am.
5 0
2 years ago
On a certain quiz with 25 questions, each correct answer is scored 4 points and each incorrect answer is scored –1 point. questi
Katarina [22]

x correct answers


23 - x incorrect answers


82 = 4x + -1(23 - x) = 3x - 23


105 = 5 x


x = 105/5 = 21


21 right 2 wrong 2 unanswered


Check: 4(21) -1(2) = 84-2=82 good



7 0
3 years ago
Read 2 more answers
Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last
lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55

The degrees of freedom are df=599

df=n-1=600-1=599

The P-value for this test statistic is:

P-value=P(t

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is  t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0
3 years ago
Asap and brainliest<br> 19) Solve. x + 4 = 10 A) x = 2 B) x = 3 C) x = 6 D) x = 14
san4es73 [151]
The answer is d. x=14.
6 0
3 years ago
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16 3/7−4 3/4 Please solve in mixed number form
nadya68 [22]

Answer: Heyaa!!

327/28

Step-by-step explanation:

<u>Convert the mixed numbers to improper fractions, then find the LCD and combine.</u>

Hopefully this helps <em>you !</em>

- Matthew  :))

8 0
2 years ago
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