All categories

3 years ago

Hamid has gained weight.

Mathematics

1 answer:

Tanya [424]3 years ago

8 0

Answer:

So, his normal weight is <u>80 kg</u>

Step-by-step explanation:

Let his normal weight be x

If the normal weight increasing percent now weighs

100 10 110

x 10 88

Now we can get a equation as,

$100 = 110\\x = 88$

We can write this as a fraction

$\frac{100}{x} = \frac{110}{88}$

You can use cross multiplication to solve this

$88*100=110x\\8800=110x\\\frac{8800}{110} =\frac{110x}{110} \\80kg = x$

Hope this helps you.

Let me know if you have any other questions :-)

You might be interested in

For this question I need to find 80% of 20

Ipatiy [6.2K]

Answer:

Step-by-step explanation:

.8 x 20 =16

7 0

3 years ago

For the following hypothesis test, H0: μ ≥ 150Ha: μ < 150the test statistic a. must be positive. b. must be negative. c. can

Virty [35]

Answer:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

And since we are conducting a lower unilateral lest then we expected that the value for the statistic would be negative. And the correct answer would be:

b. must be negative

Step-by-step explanation:

Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =150$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is lower than 150, the system of hypothesis would be:

Null hypothesis: $\mu \geq 150$

Alternative hypothesis: $\mu < 150$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

And since we are conducting a lower unilateral lest then we expected that the value for the statistic would be negative. And the correct answer would be:

b. must be negative

7 0

4 years ago

Use the interactive spinner to simulate who among Elisondra and her three friends will order first during 40 restaurant visits.

Marina CMI [18]

Answer:

0.35

Step-by-step explanation:

5 0

4 years ago

Read 3 more answers

Why is it unimportant to have the same powers of 10 we multiplying and dividing in Scientific Notation

nikdorinn [45]

We only need to worry about the exponents

6 0

3 years ago

Read 2 more answers

Complete parts (a) through (c) below.

Yuki888 [10]

Answer:

a) $t_{crit}= 1.34$

b) $t_{crit}=-2.539$

c) $t_{crit}=\pm 2.228$

Step-by-step explanation:

Part a

The significance level given is $\alpha=0.1$ and the degrees of freedom are given by:

$df = n-1= 15$

Since we are conducting a right tailed test we need to find a critical value on the t distirbution who accumulates 0.1 of the area in the right and we got:

$t_{crit}= 1.34$

Part b

The significance level given is $\alpha=0.01$ and the degrees of freedom are given by:

$df = n-1= 20-1=19$

Since we are conducting a left tailed test we need to find a critical value on the t distirbution who accumulates 0.01 of the area in the left and we got:

$t_{crit}=-2.539$

Part c

The significance level given is $\alpha=0.05$ and $\alpha/2 =0.025$ and the degrees of freedom are given by:

$df = n-1= 11-1=10$

Since we are conducting a two tailed test we need to find a critical value on the t distirbution who accumulates 0.025 of the area on each tail and we got:

$t_{crit}=\pm 2.228$

4 0

3 years ago