All categories

3 years ago

HELP PLEASEEEEEEE it’d be rlly appreciated

Mathematics

2 answers:

Alex_Xolod [135]3 years ago

4 0

Answer: they have

et n = number of nickels

Let d = number of dimes

given:

(1) +n+%2B+d+=+45+

(2) +5n+%2B+10d+=+360+

-------------------

Multiply both sides of (1) by 5

and subtract (1) from (2)

(2) +5n+%2B+10d+=+360+

(1) +-5n+-+5d+=+-225+

+5d+=+135+

+d+=+27+

and, since

(1) +n+%2B+d+=+45+

(1) +n+%2B+27+=+45+

(1) +n+=+45+-+27+

(1) +n+=+18+

She had 18 nickels and 27 dimes

check:

(2) +5%2A18+%2B+10%2A27+=+360+

(2) +90+%2B+270+=+360+

(2) +360+=+360+

Step-by-step explanation:

professor190 [17]3 years ago

3 0

Answer:18 nickels 27 dimes

You might be interested in

Jackson is going to have 5 four wheelers at his birthday party. Each four wheeler uses 4 1/3 gallons of gas how many gallons of

Lemur [1.5K]

Answer:

20 2/3

Step-by-step explanation:

i think thats what it is

5 0

3 years ago

Can anyone please help me will mark brainliest

Andrei [34K]

Answer: the price is $19.60

Step-by-step explanation:

6 0

3 years ago

A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

8 0

3 years ago

MMA-A Unit 3 packet answer key? Lmk

posledela

Answer:

yea its right here

Step-by-step explanation: sike

5 0

3 years ago

The output is less than the input

lakkis [162]

We r learning the energy unit too! what's ur question?

3 0

3 years ago