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Andrej [43]
2 years ago
9

Write the inequality!! Connie’s dog Fido weighs 35 pounds. Her vet placed Fido on a diet. What inequality can you write to find

the average number of pounds Fido must lose monthly to reach a healthier weight of 28 pounds within 6 months?
Mathematics
1 answer:
Butoxors [25]2 years ago
6 0

Answer:   \frac{35-28}{x} \leq 6

Step-by-step explanation:

Let x be the average number of pounds Fido must loss.

Since, the initial weight of Fido is 35 pounds.

And, After losing the weight, the new weight of Fido in pounds = 28 pounds.

Then the time taken for losing the weight

= \frac{\text{ The weight it losses}}{\text{ Average of losing weight per month}}

= \frac{35-28}{x}

According to the question,  it must lose weight within 6 months,

Thus,   \frac{35-28}{x}\leq 6

Which is the required inequality to find the average number of pounds per month.

By solving it we, get,    x\geq \frac{7}{6}

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What is the median for this data set?
Sever21 [200]
The meridian for this set is 3.005 (option C)

Note ➡The word Median is lime the "average" for all of the values of the particular problem
8 0
3 years ago
Line m is parallel to line p. m ∠ HEF = 39º and m ∠ IGF = 13º. Find the m ∠ EFG. Explain in detail how you know you are correct.
svetlana [45]

Answer:

The measure of ∠EFG is 52°

Step-by-step explanation:

Given line m is parallel to line p. m∠HEF = 39º and m∠IGF = 13º.we have to find m∠EFG.

In ΔJFG,

By angle sum property of triangle, which states that sum of all angles of triangle is 180°

m∠FJG+m∠JGF+m∠JFG=180°

⇒ 39°+13°+m∠JFG=180°

⇒ m∠JFG=180°-39°-13°=128°

As JFE is a straight line ∴ ∠JFG and ∠EFG forms linear pair

⇒ m∠JFG+m∠EFG=180°

⇒ 128°+m∠EFG=180°

⇒ m∠EFG=52°

The measure of ∠EFG is 52°

6 0
3 years ago
A Cell Phone company sells cellular phones and airtime in a State. At a recent meeting, the marketing manager states that the av
GarryVolchara [31]

Answer:

The null hypothesis is rejected.

There is enough evidence to support the claim that the average age of the customers differs from 40 years.

The sample does not support the manager claim (the average age seems to differ from 40 years).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The manager claims that the average age of customers is 40 years. As this is an equality, we will test if the average age differs from 40. If the null hypothesis failed to be rejected, the claim of the manager is right.

Then, the claim is that the average age of the customers differs from 40 years.

Then, the null and alternative hypothesis are:

H_0: \mu=40\\\\H_a:\mu\neq 40

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=38.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=7.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{7}{\sqrt{50}}=0.99

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{38-40}{0.99}=\dfrac{-2}{0.99}=-2.02

The degrees of freedom for this sample size are:

df=n-1=50-1=49

This test is a two-tailed test, with 49 degrees of freedom and t=-2.02, so the P-value for this test is calculated as (using a t-table):

P-value=2\cdot P(t

As the P-value (0.049) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average age of the customers differs from 40 years.

3 0
3 years ago
there are 112 male and 78 females working at the hospital. what percentage of the hospital workforce is female
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2 years ago
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Round 1395 to the nearest hundred
algol [13]

Answer:

1400

Step-by-step explanation:

It is closer to 1400 than 1300

5 0
3 years ago
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