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givi [52]
3 years ago
11

Evaluate the following expression given that z=8 2z-4

Mathematics
1 answer:
PtichkaEL [24]3 years ago
5 0

Answer:

12

Step-by-step explanation:

2 times 8 is 16. 16 minus 4 is 12.

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Find the distance from the base of the ladder to the house.
True [87]

Answer:

tfeybgchhimliehbu uhuijo;k,l

Step-by-step explanation:

1.vuybcewouinjhi

2.dfbvfdsa

7 0
3 years ago
Una señora compro medio kilo de pescado y 450 gramos de carne de res. ¿Cuánto le falta para completar el kilo?
matrenka [14]

Answer:

LE FALTA 50 gr

Step-by-step explanation:

PESCADO MEDIO KG= 500 gr

CARNE DE RES= 450 gr

500+450=950 gr

1000 - 950= 50 gr

8 0
2 years ago
PLEASE HELP WILL GIVE BRAINLIEST ANSWER!!
andreyandreev [35.5K]
Speed = Distance / Time

Distance = 5t^4 - 10t^2+6

Time = t+2

Now:

 5t^4 - 10t^2 + 6         | t+2
-5t^4 -10t^3                 5t^3 - 10t^2 + 10t - 20
 \\\\\ -10t^3-10t^2+6
       +10t^3+20t^2
        \\\\\\\ 10t^2+6
                -10t^2-20t
                 \\\\\ -20t+6
                      +20t+40
                       \\\\\ 46

<span>C. 5t^3-10t^2+10t-20+(46)/(t+2))</span>
3 0
3 years ago
An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than
Sedbober [7]

Testing the hypothesis, it is found that:

a)

The null hypothesis is: H_0: \mu \leq 10

The alternative hypothesis is: H_1: \mu > 10

b)

The critical value is: t_c = 1.74

The decision rule is:

  • If t < 1.74, we <u>do not reject</u> the null hypothesis.
  • If t > 1.74, we <u>reject</u> the null hypothesis.

c)

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of <u>at most 10 pounds</u>, that is:

H_0: \mu \leq 10

At the alternative hypothesis, it is tested if the mean loss is of <u>more than 10 pounds</u>, that is:

H_1: \mu > 10

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a <u>significance level of 0.05</u> and 18 - 1 = <u>17 df.</u>

Hence, using a calculator for the t-distribution, the critical value is: t_c = 1.74.

Hence, the decision rule is:

  • If t < 1.74, we <u>do not reject</u> the null hypothesis.
  • If t > 1.74, we <u>reject</u> the null hypothesis.

Item c:

We have the <u>standard deviation for the sample</u>, hence the t-distribution is used. The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

For this problem, we have that:

\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18

Thus, the value of the test statistic is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}

t = 1.41

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0
3 years ago
Hi plz answer this I'll give u more points if u do
Brrunno [24]

Answer:

6 feet in the distance

Step-by-step explanation:

6 0
2 years ago
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