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2 years ago

PLEASE HELP I NEED TO PASS MATH!

Mathematics

2 answers:

fgiga [73]2 years ago

6 0

Answer:

a translation moves a thing up and down or left and right.

Step-by-step explanation:

so true?

Ulleksa [173]2 years ago

3 0

Answer:

True

Step-by-step explanation:

A translation means you are moving a figure in some direction(s). This can mean both vertically and horizontally.

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Please help! Its a test that’s due today.

Eduardwww [97]

Answer:

it's b

Step-by-step explanation:

the denimonater is the root of the radical and the power

Hope this helps :)

3 0

3 years ago

Read 2 more answers

Choose the correct discriminant, then

irina [24]

Answer:

Step-by-step explanation:

Order of Operations; PEMDAS

Parentheses

Exponents

Multiplication & Division (Left to Right) (ex: 10x2÷5) You would do whatever comes left for Multiplication or Division.

Addition & Subtraction (Left to Right) (ex: 10-2+5) You would do whatever comes left for Addition & Subtraction.

3x+6 =

(9+6=15)

(6x2=12)

(12-15=<u>3</u>)

7 0

3 years ago

A bag contains 20 slips of paper Individually numbered. If a slip of paper and a letter in the word PENNSYLVANIA is chosen at ra

Soloha48 [4]

Answer:

$P(M4\ n\ V) = 2.083\%$

Step-by-step explanation:

Given

Paper = 20 slips

Word: PENNSYLVANIA

Required

Determine P(Multiple of 4 and V)

The sample size of the 20 slips is:

$n(S) = 20$

The outcomes of multiples of 4 is:

$M4= \{4,8,12,16,20\}$

$n(M4) = 5$

So, the probability of multiples of 4 is:

$Pr(M4) = \frac{5}{20}$

$Pr(M4) = \frac{1}{4}$

The sample size of PENNSYLVANIA is:

$n(S) = 12$

The outcome of V is:

$n(V) = 1$

So, the probability of V is:

$P(V) = \frac{1}{12}$

So, the required probability is: P(Multiple of 4 and V)

$P(M4\ n\ V) = P(M4) * P(V)$

$P(M4\ n\ V) = \frac{1}{4} * \frac{1}{12}$

$P(M4\ n\ V) = \frac{1}{48}$

$P(M4\ n\ V) = 0.02083$

Express as percentage

$P(M4\ n\ V) = 0.02083 * 100\%$

$P(M4\ n\ V) = 2.083\%$

4 0

3 years ago

The daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6. What ar

Degger [83]

Answer:

The minimum value of the bill that is greater than 95% of the bills is $37.87.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 28, \sigma = 6$

What are the minimum value of the bill that is greater than 95% of the bills?

This is the 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 28}{6}$

$X - 28 = 6*1.645$

$X = 37.87$

The minimum value of the bill that is greater than 95% of the bills is $37.87.

4 0

3 years ago

Use the sample information 11formula13.mml = 37, σ = 5, n = 15 to calculate the following confidence intervals for μ assuming th

bija089 [108]

Answer & Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

σ= standard deviation. In this case 5

mean= 37

n= number of observations. In this case, 15

(a)

Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645

Then, the confidence interval (90%):

I 90%(μ)= 37+- [1.645*(5/sqrt(15))]

I 90%(μ)= 37+- [2.1236]

I 90%(μ)= [37-2.1236;37+2.1236]

I 90%(μ)= [34.8764;39.1236]

(b)

Z(alpha/2)= Z(2.5%)= 1.96

Then, the confidence interval (90%):

I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]

I 95%(μ)= 37+- [2.5303]

I 95%(μ)= [37-2.5303;37+2.5303]

I 95%(μ)= [34.4697;39.5203]

(c)

Z(alpha/2)= Z(0.5%)= 2.5758

Then, the confidence interval (90%):

I 99%(μ)= 37+- [2.5758*(5/sqrt(15))

I 99%(μ)= 37+- [3.3253]

I 99%(μ)= [37-3.3253;37+3.3253]

I 99%(μ)= [33.6747;39.3253]

(d)

C. The interval gets wider as the confidence level increases.

8 0

3 years ago