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Cloud [144]
3 years ago
5

Ricardo's dog weighs 6 times as much as his cat.The total weight of his two pets is 98 pounds. How much does Ricardo's dog weigh

?
Mathematics
2 answers:
Nina [5.8K]3 years ago
8 0
98 pounds : 7 = 14 pounds
14 pounds x 6 = 84 pounds
Ricardo's dog weighs 84 pounds
OverLord2011 [107]3 years ago
4 0
Ricardo's cat weighs x pounds.
His dog weighs 6 times as much as his cat, so his dog weighs 6x pounds.
The total weight of the pets is 98 pounds.

x+6x=98 \\
7x=98 \\
x=\frac{98}{7} \\
x=14 \\ \\
6x=6 \times 14=84

Ricardo's dog weighs 84 pounds.
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What is the sum of this expression??<br><br> 3 3/7 + 1 6/7 = ???
Anna11 [10]

Answer:

5 2/7

Step-by-step explanation:

3 + 1 = 4

3/7 + 6/7 = 9/7 = 1 2/7

1 2/7 + 4 = 5 2/7

3 0
3 years ago
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ANTONII [103]
Multiply 400 times 0.70 and that’s how you get the % (in this case 280)
5 0
3 years ago
A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled
klasskru [66]

Answer:

The standard error of the mean is 4.5.

Step-by-step explanation:

As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5

The standard deviation of the sample is calculated as:

s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8

Then, we can calculate the standard error of the mean as:

\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5

6 0
3 years ago
Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

Z= \frac{(X[bar]-Mu)}{\frac{Sigma}{\sqrt{n}}  } ~N(0;1)

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

6 0
3 years ago
Y=x+2<br> and then you have x=-3 <br><br><br> You also have to use transitive property
tresset_1 [31]
If y = x+2 and x = -3 then we can say that y = -3+2 after replacing the 'x' with -3

From here, we just add -3 and 2 to get -1 which is why y = -3+2 turns into y = -1

So together x = -3 and y = -1

They pair up to form (x,y) = (-3,-1)
8 0
3 years ago
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