2 years ago

A doctor estimates that a particular patient is losing bone density at a rate of 3% annually. The patient currently has a

Mathematics

1 answer:

MA_775_DIABLO [31]2 years ago

6 0

it is 1500 and .97

Hope this helps! Have a blessed day!

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lbvjy [14]

Answer:

$4 {x}^{2} - 2x + x = 5 {x}^{2} + 2x \\ 4 {x}^{2} - x = 5 {x}^{2} + 2x \\ 4 {x}^{2} - 5 {x}^{2} = 2x + x \\ - 1 {x}^{2} = 3x \\ - x = 3 \\ x = - 3 \\ thank \: you$

5 0

2 years ago

Read 2 more answers

i need a data set for a project and i cant think of 2 things to compare please help quick will give brainliest please give me 2

Zinaida [17]

Answer:

ok so here... 1:people and animals 2:math and science 3:shoes and sandals 4:cats and dogs thats all i could have thinked of.

Step-by-step explanation:

5 0

3 years ago

896, -448, 224, -112 next in the sequence

faltersainse [42]

56, you are dividing by -2 each time

4 0

3 years ago

To Mark or decided to purchase a new DVD player on an installment loan DVD player was $295 Tim agreed to pay $31 per month for 1

Verizon [17]

Answer: ⇒ The finance charge = $77

Step-by-step explanation:

Given: The cost price of DVD player= $295

Since, Tim agreed to pay $31 per month for 12 months .

Therefore, the total amount paid by Tim= $31\times12=\$372$

The finance charge = $\text{Amount paid by Tim - Cost price of DVD}$

⇒ The finance charge = $\$372-\ $295$

⇒ The finance charge =$77

5 0

3 years ago

Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypot

Nina [5.8K]

Answer: a. 0.1031; fail to reject the null hypothesis

Step-by-step explanation:

Given: Significance level : $\alpha=0.05$

The test statistic in a two-tailed test is z = -1.63.

The P-value for two-tailed test : $2P(Z>|z|)=2P(Z>|-1.63|)=0.1031$ [By p-value table]

Since, 0.1031 > 0.05

i.e. p-value > $\alpha$

So, we fail to reject the null hypothesis. [When p< $\alpha$ then we reject null hypothesis ]

So, the correct option is a. 0.1031; fail to reject the null hypothesis.

8 0

2 years ago